We need answer in Chinese likely. Need solve math problem. Problem: Find smallest positive integer n such that binary decimal representation of 1/n contains binary representations of each number 1,2,...,1990. Need parse: "1/n 的二进制小数表示中能够包含 1,2,...,1990 各个数字的二进制表示" Means the binary expansion of 1/n (after binary point) contains as substrings the binary representations of integers 1 through 1990. Need smallest positive integer n. No programming brute force, no internet. Need solve mathematically.

We need think deeply. We need find minimal n such that the infinite binary expansion of 1/n contains all binary strings that are representations of numbers 1..1990 (no leading zeros). Since binary representation of 1 starts with "1", 2 "10", 3 "11", ..., up to 1990. Note if a string contains all length-L binary strings with no leading zero? Let's analyze.

We need the binary fractional expansion of 1/n. For rational 1/n, binary expansion is eventually periodic. If n has factor 2^a times odd m, expansion = preperiod length a then repeating period = ord_m(2) (if m>1). It contains a given finite binary string iff string occurs in periodic part (or preperiod). We can choose n. Need smallest integer n, not smallest denominator in terms of period? We need find minimal n such that all required substrings occur somewhere in the complete expansion (integer part 0? 1/n<1 for n>1; for n=1, 1.000... maybe contains? no). Smallest positive integer n likely odd? Since factor 2 only shifts expansion and adds leading zeros after point before repeating part; cannot help create nonzero patterns? Need prove minimal n odd and maybe n = 2047? Let's explore.

We need contain binary representations of numbers 1..1990. The largest is 1990. Binary length of 1990: 2^10=1024, 2^11=2048, so length 11. Specifically 1990 decimal = 1024+512+256+128+64+? 1990-1984=6, so bits 11111000110? Let's compute: 1990 = 1024+512=1536; +256=1792; +128=1920; +64=1984; remainder 6=4+2. Bits from 10 to 0: 1 1 1 1 1 0 0 0 1 1 0 = 11111000110 (yes). Required strings include all numbers with lengths 1 through 11 that start with 1. Does containing all numbers 1..1990 imply containing all binary strings of lengths up to 10? Let's see: For lengths 1..10, every binary string starting with 1 of length L corresponds to integer from 2^{L-1} to 2^L-1. For L<=10, max 1023 <=1990, so all length-L strings starting with 1 are included. Since any binary string (possibly with leading zeros) appears as substring of some longer string starting with 1? We only require strings without leading zeros for numbers. But if expansion contains all length-11? Not all length-11 starting with 1, only those integers <=1990, i.e. from 10000000000 (1024) to 11111000110 (1990). That is length-11 strings s starting with 1 and value <=1990. Since threshold high (1990/2047 ~0.972), missing length-11 strings are those >11111000110: from 11111000111 to 11111111111. Count 2048-1-1990? Let's list missing length-11 binary strings starting with 1 with value >1990: values 1991..2047 = 57 strings. They are not required directly. But they may be required as substrings of smaller number representations? No, required representations are exact strings for each integer. But to contain all required strings, due to substring inclusion, if a longer required string contains a shorter non-required string that's irrelevant. Need find a sequence (binary expansion of 1/n) containing a set S of words: all binary words starting with 1 of lengths 1..10 (that's all words with no leading zeros length <=10) plus length-11 words from 1024 to 1990.

Important: If a sequence contains all words of length 10 starting with 1, does it automatically contain all words length <=10 starting with 1? Not necessarily; e.g. contains "1000000000" (512) contains substring "1" and maybe many zeros but not "11". But because it contains all length-10 words starting with 1, any shorter word w starting with 1 can be extended to a length-10 word starting with w (append arbitrary bits) whose value <=1023? If w length l<=10, choose extension to length 10. That word starts with 1 and value between 512 and 1023, included. Occurrence of the longer word contains w as prefix. Yes. So requiring all numbers 512..1023 (length 10) suffices for all <=1023? For any number <512 with binary length l<10, extend its representation on the right with zeros to length 10; value = x2^{10-l}, which is between? If x>=1, value >=2^{10-l}; for l<=9, value could be as low as 512? If x=1 (binary "1"), extend with nine zeros -> 512, included. If x has length l, x2^{10-l} is a length-10 number starting with same prefix, included (<=1023). Occurrence contains representation of x as prefix. Good. So the core requirement reduces to containing all length-10 binary strings starting with 1 (i.e. all 512 words of length 10 with first bit 1) AND containing the 967? length-11 words from 1024 to 1990. Wait if sequence contains all length-10 words starting with 1, plus some length-11 words. For length-11 required words <=1990, many may occur by extending length-10? No occurrence of a length-10 word does not guarantee length-11 with needed next bit. Need specifically ensure length-11 words in [1024,1990]. But maybe containing all length-11 words starting with 1 up to threshold T is implied by containing all length-10 words? Not all; to have a given length-11 word v = u b where u length 10 starts with 1. Since u occurs somewhere, but occurrence may be followed by wrong bit. However because all length-10 words occur in a periodic sequence maybe de Bruijn properties.

Maybe minimal n relates to a de Bruijn sequence of order 11 or 10. For a binary periodic sequence of period N to contain all binary strings of length k (with no leading zero restriction maybe all length-k strings), minimal period is 2^k (de Bruijn). But we only require strings of length 11 starting with 1 and <= threshold, plus via prefix all shorter. Actually to contain all length-10 words starting with 1, since any occurrence of a length-10 word starting with 1 can be followed by arbitrary; but what about length-10 words starting with 0? Not required. However the cyclic sequence of period p for an odd denominator m: the repeating part has length r=ord_m(2). We need choose m (odd) minimal such that the purely periodic binary expansion of 1/m contains required words. Also n could have factor 2; minimal n likely odd because multiplying by 2 increases n and just shifts right (adds a zero after point) not adding needed nonzero patterns? Need prove.

For rational 1/n, binary expansion's eventual period depends on odd part m. Let n=2^a m with m odd. The fractional bits: preperiod a zeros? Let's compute: 1/(2^a m) = 2^{-a} * (1/m). The binary expansion of 1/m is 0.\overline{p} (purely periodic) for odd m>1; shifting right by a gives a zeros after point then period p. If m=1, expansion terminates (1/2^a = 0.000...1?) Actually 1/(2^a) = 0.0...01 (then zeros) not containing. For m>1, repeating part same as 1/m shifted; any nonzero word occurring in shifted expansion occurs in period p (unless it overlaps the preperiod zeros and period; but words starting with 1 cannot start in preperiod zeros except maybe at the last preperiod? If a>=1, the first bit of repeating period could combine with preceding zeros? Required words all start with 1, so an occurrence could start at the first bit of period or later; cannot start in preperiod because preperiod bits are zero. If a=0 no preperiod. If a>0, there are a zeros after point; a required word starting with 1 cannot include those zeros at beginning. It could start at position <=a? bit is 0 no. Could start at position a+1 first periodic bit. So all required occurrences are in periodic part. Thus factor 2^a only increases n without changing available required words. For minimal n, a=0. Also m=1 not possible. So n odd >1.

Thus problem reduces: find smallest odd positive integer n such that the repetend of 1/n (length r=ord_n(2)) contains all required binary words. Since the expansion of 1/n is purely periodic with period r. But note the repeating block for 1/n is not arbitrary: it is the binary representation of floor(2^r/n) maybe with leading zeros to length r, and the sequence is the rotation corresponding to fractional parts 2^k/n. More precisely for odd n, let r=ord_n(2). The period bits are the r-bit binary representation of A=(2^r-1)/n (since 1/n = A/(2^r-1) = 0.\overline{A_bits}). Indeed if n divides 2^r-1 minimal r, 1/n = q/(2^r-1) with q=(2^r-1)/n, and the repeating block is q written in r bits (allow leading zeros). Need find smallest odd n such that the r-bit block B = binary(q, r) (cyclic infinite repetition) contains all required words. Here r=ord_n(2), and n q = 2^r-1. Since q integer. Because n minimal maybe r about 11? Need contain length-11 patterns, so r must be at least 11? A periodic sequence of period r contains at most r distinct substrings of length 11 (cyclic). We need at least all length-10 starting with 1 (512 substrings) as length-10. For length 10, a period-r cyclic sequence has at most r distinct length-10 substrings. To have 512 required distinct length-10 substrings, need r >=512. Wait yes! For cyclic binary sequence of period r, number of length-10 substrings (counted with start positions) is r (if r>=10; distinct ≤r). We require 512 distinct length-10 words (all starting with 1). Thus r ≥ 512. Is that possible? r=ord_n(2) must be ≥512. Minimal odd n with ord_n(2)≥512 and with quotient block containing required set. Lower bound n? Since n divides 2^r-1 and ord_n(2)=r. If r≥512, then n ≤ 2^r-1 but we want smallest n. Could there be small n with huge multiplicative order? For n prime, order can be n-1. Need n minimal such that ord_n(2) ≥512. That would be smallest prime p with 2 a primitive root mod p and p-1 ≥512? The smallest prime p>512 for which ord_p(2)=p-1 maybe. If p=521? 521 prime? 521 =? 521= ? 521/7=74.4; 521=865+1; divisibility: by 13? 1340=520+? no; 521 prime? sqrt=22.8; test 3 no (5+2+1=8),5 no,7 774=518 rem3,11 1147=517 rem4,13 1340=520 rem1,1730=510 rem11,1927=513 rem8,23>sqrt. prime. Is ord_521(2)=520? Need check. 520=8513. Need 2^{520/q} not 1 for q=2,5,13. 2^260 mod521 =? Legendre symbol (2/p)=(-1)^{(p^2-1)/8}. p=521 ≡1 mod8? 521 mod8=1, so (2/p)=1, so 2^260=1 mod p. Thus order divides 260, not ≥512. Next prime with p ≡ ±3 mod8 for 2 to be nonsquare. We need order≥512. Since ord divides p-1, p-1≥512 => p≥513. Smallest prime p≥513 with order≥512. p-1 must be ≥512; for prime p, ord_p(2) divides p-1. If p<521 no. p=521 order divides 260 due p≡1 mod8? Actually order divides (p-1)/2 if 2 square. So max 260 <512. Next primes: 523 (prime? sqrt22.8; mod8=3, so 2 nonsquare; order does not divide (p-1)/2=261? Wait if 2 nonsquare, 2^{(p-1)/2}=-1, so order has full 2-adic factor but may divide 522=23^229. Order could be 522 if primitive; need check divisors. For order≥512 and divides 522, the only divisor ≥512 is 522 itself (since proper divisors ≤261? Divisors of 522: 1,2,3,6,9,18,29,58,87,174,261,522. Yes only 522 ≥512). So need ord_523(2)=522. Is 2 primitive mod523? Need check 2^{522/q} ≠1 for q=2,3,29. Since p≡3 mod8, 2^{261}=-1 ≠1 good. Need 2^{174} ≠1 (q=3) and 2^{18} ≠1 (q=29). 2^18=262144 mod523. Compute 523501=262? 523500=261500; remainder644; +523=262023 rem121; so 2^18 ≡121 mod523, not 1. 2^174? Harder. Could compute via math manually? Maybe. But maybe n not prime and smaller n with ord≥512? n must be at least? If ord_n(2)=r≥512, then r divides λ(n) ≤ φ(n) < n for n>2. Actually ord_n(2) ≤ λ(n) ≤ φ(n) ≤ n-1 (for n>2). Therefore n ≥ r+1 ≥513. Since n odd, n≥515? Wait if ord_n(2)≥512 then n > ord, so n ≥513. But n odd => n≥515. However if n composite, λ(n) can be ≥512 with n maybe less than 513? No λ(n)≤φ(n)≤n-1, so n≥513. Odd n≥515. But for n=515=5103, λ=lcm(4,102)=204? <512. n=517=1147 λ=lcm(10,46)=230. n=519=3173 λ=lcm(2,172)=172. n=521 prime but order≤260. n=523 prime possible. So lower bound n≥523? Need check n=515,517,519,521 cannot have ord≥512 (λ<512 for composites; p=521 λ=520 but 2 square order≤260). What about n=513 odd? 513=3^319? φ=513(2/3)(18/19)=324? <512. n=515 φ=408; n=517 φ=460; n=519 φ=344; n=521 φ=520 but ord≤260; n=523 candidate. Good. Need prove minimal n=523 if 1/523 expansion contains required patterns. Does period r=522 (if primitive) contain all length-10 strings starting with 1? A cyclic sequence of length 522 has only 522 length-10 substrings. Need contain 512 specific ones. Possible but not guaranteed. The repeating block for 1/523 is q=(2^522-1)/523, enormous. We cannot brute force. Need derive mathematically. Hmm maybe minimal n is not based on r≥512? Let's reconsider. The binary expansion of 1/n with n=523 has period 522 if 2 primitive. A de Bruijn sequence of order 10 has period 1024 and contains all 1024 length-10 binary strings. But we only need the 512 length-10 strings starting with 1. A period of 522 can contain at most 522 distinct length-10 strings; containing all 512 that start with 1 would mean almost every length-10 substring starts with 1. In any binary cyclic sequence of length r, number of length-10 substrings starting with 1 equals number of positions i with bit s_i=1 (since substring of length 10 starts with bit s_i), provided r≥10. Exactly the Hamming weight of the period block! To contain all 512 distinct length-10 words starting with 1, we need at least 512 substrings starting with 1, so weight ≥512. But the repeating block of 1/n, q=(2^r-1)/n, may have weight around r/2 if n small? For n=523, r=522, q≈2^522/523 ≈ 2^522/ (≈2^9.03) = 2^512.97, so q is about 513 bits? Wait q has r bits (522) with leading zeros? q ≈ (2^522)/523 ≈ 2^522 / 2^9.03 = 2^512.97, so as a 522-bit block it has about 9 leading zeros? Let's compute: q < 2^513 maybe, because (2^522-1)/523 < 2^522/512 = 2^513. Since 523>512, q<2^513. Also q > (2^522-1)/523 > about 2^512.99? Need determine bit length of q. If q <2^513, then the 522-bit repeating block begins with at least 9 zeros. In fact the binary expansion of 1/n for n>512 begins with 9 zeros after point? Since 1/523 < 1/512 = 2^-9, but >1/1024=2^-10, so first 9 fractional bits are zero and the 10th bit is 1. For n=523 (between 513 and 1023), 1/n is between 2^-10 and 2^-9, so bits positions 1..9 are 0, position 10 =1? Let's check n<1024 =>1/n>2^-10; n>512=>1/n<2^-9. Binary: bit9 corresponds 2^-9=1/512. Since 1/n<1/512, bit9=0. bit10=1/1024; since 1/n>1/1024, but to decide bit10 after nine zeros, value >=2^-10? For n<1024, 1/n >1/1024; if all first 9 bits zero, the max with bit10=0 is <2^-10 (actually can approach 2^-10 from below with later ones), so bit10 must be 1? For n=1023, 1/1023 ≈0.000000000111111... binary bits first nine? 1/1023 = 1/(1024-1)=2^-10/(1-2^-10)=2^-10+2^-20+... so bits: positions1-9 zero, position10=1, positions11-19 zero? yes. For n=523, first one at position10. Thus the repeating block has leading zeros; as cyclic sequence, there are runs of zeros. The Hamming weight maybe? q≈2^513 so bit length 513? Then among r=522 block there are 9 leading zeros and rest maybe ~? Weight of q maybe not controllable. Number of length-10 substrings starting with 1 = weight of block (cyclic). Need at least 512 ones. But if q has only 513 bits, can it have Hamming weight ≥512? That would mean q = 2^513 - 1 - small? Let's examine. q=(2^522-1)/n. For n just above 512, q just below 2^513. If n=513? q≈(2^r? not ord) not relevant. For n=523 with r=522, q is floor? exactly (2^522-1)/523. Approx q = 2^522/523. Compare to 2^513: q/2^513 = 2^9/523 =512/523 ≈0.978967. So q ≈0.9789672^513. In binary, q is a 513-bit number (since >2^512) with leading bits (relative to 513 bits) equal to about 0.978967, i.e. top 9? Let's express q as 513-bit string (bits b512...b0). It begins with 1 (since >2^512). The value deficit from 2^513 is d=2^513 - q ≈ (1 -512/523)2^513 = (11/523)2^513 ≈0.02103252^513 ≈? That's about 2^513 /47.55 ≈2^507.43. So q = 2^513 - d where d≈2^507.43. Binary of q: since d <2^508 maybe, q = (2^513 -1) - (d-1). 2^513-1 is 513 ones. Subtracting d-1 (~508-bit) flips only lower ~509 bits; the top 5 bits remain ones? More precisely if d<2^508, then d-1 <2^508, so q = (2^513-1) - (d-1) has top 513-508=5 bits all ones, and lower 508 bits are complement of d-1. Then Hamming weight of q would be 5 + (508 - popcount(d-1)) ≈513 - popcount(d-1) ≈513-254=259, not ≥512. Wait I made a mistake: If q≈0.9792^513, binary starts 111110... maybe top bits: 0.979 in binary = 111110101...? Let's compute. 512/523 = 1 -11/523. 11/523≈0.0210325. Binary of that: 0.000001010...? 2^-6=0.015625, remainder .0054; 2^-8=.003906; etc. So 512/523 ≈0.1111101011... (not 0.978 with many leading ones? 0.979 decimal binary is 0.111110101... yes top six bits 111110). I incorrectly set q=2^513 - d with d≈0.0212^513=2^507.43. If d≈2^507.4 (<2^508), then q = 2^513 - d indeed has top 5 bits? Let's test with small example: 2^13 - d where d=0.0218192≈172 (<256=2^8). 8192-172=8020. Binary 8020 = 8192-172 = 0b1111101010100? 13 bits: top bits? 8192 (bit12) minus 172 -> 8020 = 4096+2048+1024+512+256+64+16+4 = bits 12..8? 8020/8192=0.979; binary 1111101010100 (13 bits) has zeros at bit7? Let's see 2^13-1=8191 (13 ones). Subtract d-1=171 (8-bit) from 8191 yields top 5 bits (bits12..8) remain 11111? 8191-171=8020 binary 1111101010100. Top five bits bits12-8 = 11111, bit7=0, lower bits complement. Hamming weight =5 + (8 - popcount(171)) + lower? Wait d-1=171 (<2^8). 8191 (13 ones) -171 = top (13-8=5) ones plus (2^8-1 -171)=84 in lower 8 bits. popcount(84)=4 (1010100). Weight=5+4=9, equals 13-popcount(171)? popcount171=4, yes weight=9. So weight ≈ r - popcount(d-1) ≈513-~4? No d-1≈2^507 not small, popcount ~253. So weight ~260. Right.

But for n just above 512, the repeating block q in r bits has a long run of leading zeros? Let's be careful: q has bit length L=floor(log2 q)+1. For n=523 and r=522, q≈2^512.97, so L=513. The r-bit block has r-L=9 leading zeros, then 513 bits of q. Its Hamming weight = popcount(q). q≈(512/523)2^513 = 2^513 - (11/523)2^513. The subtracted d ≈0.0210*2^513 ≈2^507.43. popcount(q) not simply ~260? Yes if d is random 508-bit, q = (2^513-1) - (d-1) but only if d<2^513; however d has bit length ~508, so complementing lower 508 bits yields about 5 + (508 - popcount(d-1)) ≈513 - popcount(d-1). If d-1 random 508-bit popcount≈254, popcount(q)≈259. So the block has only ~259 ones, not ≥512. But wait q as 513-bit string begins with 111110...; weight maybe ~half of 513 =256. Yes.

Then number of length-10 substrings starting with 1 = popcount(block) ≈259 <512. Impossible to contain 512 distinct length-10 words. This suggests n=523 cannot work. Need rethink lower bound: To have at least 512 length-10 substrings starting with 1, the repeating block must have Hamming weight at least 512. For 1/n with period r and quotient q=(2^r-1)/n, the r-bit block's Hamming weight is popcount(q). We need popcount(q)≥512 (and also distinct substrings). Since q=(2^r-1)/n. For small n, q is large and maybe popcount large. We want smallest n; n around? If n is small (e.g. n=3), period r=2, cannot have length 10. Need r≥512. But also q roughly (2^r)/n. If n is small relative to 2^r, q has about r - log2 n bits? Actually q is r-bit block maybe with leading zeros if n>1. For n<2^r, q>1. The r-bit representation of q may have leading zeros if q<2^{r-1}, i.e. if (2^r-1)/n <2^{r-1} => n > 2 - 2^{-(r-1)} => n≥3? For n≥3, q<2^{r-1}; so the block's leading bit is 0. In fact 1/n<1/2 so first fractional bit 0 for n>2. For n between 2^{k-1}+1 and 2^k, first k-1 bits? General: if n>2^a, first a bits zero. To have weight≥512 with period r, need enough ones. For n small, q close to (2^r)/n; if n has small value, q = (2^r-1)/n may not be integer? It is if n divides 2^r-1. For n with ord r, q is cofactor. If n is small but ord r large (possible if n is a divisor of 2^r-1 with large order; n itself ≥ r+1 as argued, so n≥513 for r≥512). Thus n cannot be small; minimal n ≥ r+1≥513. Then q≈2^r/n ≤2^r/513. The r-bit block has at least about 9 leading zeros and only about r-9 significant bits. For r=512 minimal, n≥513 but n must divide 2^{512}-1 and have ord 512. However n≥513 and n divides 2^{512}-1? If ord_n(2)=512, then n divides 2^{512}-1 and not smaller. But since n>512, and 2^{512}-1 huge, possible divisors. q=(2^{512}-1)/n < (2^{512})/513 ≈2^{502.996}; bit length ≤503, so weight≤503<512. Therefore r=512 cannot yield weight≥512 if n≥513. More generally, for given r≥512, minimal possible n with ord r is at least r+1 (since ord≤n-1), so q=(2^r-1)/n <2^r/(r+1). Then bit length of q ≤ r - floor(log2(r+1)) maybe. For r around? Need weight≥512. If n≥r+1, q<2^r/(r+1). For r=522, q<2^r/523 ≈2^{512.97}, bit length≤513; weight≤513. Wait earlier I said weight≈259, but upper bound bit length=513 permits weight up to 513 (if q=all ones in lower 513 bits). Could q be very close to 2^{513}-1? That requires n close to 2^{r-513}? Let's derive exact: For r=522, q<2^522/523. The maximum possible integer q with divisor n≥523 is floor((2^522-1)/523). We estimated q≈(512/523)2^513, which is not close to 2^513-1; it's about 0.9792^513, i.e. top bits 111110..., and because the subtracted d is large (~2^507), weight could be around half, not near 513. But the upper bound bit length 513 is not enough to rule out weight≥512. Need stronger: q ≤ floor((2^r-1)/n). For n≥r+1=523, q < 2^522/523. To have popcount≥512, q must be at least the smallest 513-bit number with popcount≥512, i.e. at least 2^{513}-1 - (something). The smallest number with bit length ≤513 and popcount≥512 is? To maximize small value with high popcount, choose lower 512 bits ones and all higher zeros: q_min_weight512 = 2^{512}-1 (popcount512) actually value ~2^512. Wait popcount≥512 does NOT require q close to 2^513; the smallest number with 512 ones is 2^{512}-1 (bits 0..511 all one), which is ~2^512. Our q≈2^512.97 >2^512, so possible. Right! I mixed. The r-bit block has 9 leading zeros, then q maybe up to 513 bits. To have weight≥512, q must have at least 512 ones among its ≤513 bits. That means q is extremely close to having all bits 1: among its 513-bit representation, at most one zero. Because bit length could be 512 or 513. If q has bit length 512 and popcount≥512, then q=2^{512}-1 exactly. If q has bit length 513 and popcount≥512, then q ≥ 2^{512} + (2^{512}-1) - (one lower bit?) More precisely q in [2^{512},2^{513}) with at most one zero among 513 bits. The smallest such q is 2^{512} + (2^{512}-2)?? Let's find: 513 bits (bit512=1 plus lower 512 bits) with at most one zero total. Since bit512 must be 1 (bit length 513), among lower 512 bits at most one zero. Smallest value occurs with the zero at the highest possible lower bit? To minimize value with at most one zero among lower bits, set the single zero at the most significant lower bit (bit511) and all other lower bits 1: q=2^{512} + (2^{512}-1 -2^{511}) = 2^{512}+2^{511}-1 = 32^{511}-1 ≈1.52^{512}=0.752^{513}. If no zero, q=2^{513}-1. So threshold for popcount≥512 with bit length 513 is q ≥ 32^{511}-1 = (3/4)2^{513}-1. Our q≈0.97892^{513} is well above, so popcount could be ≥512? Wait if q=0.9792^513, binary begins 111110..., that has a zero at bit? Let's compute relative to 513 bits: q/(2^513)=512/523=0.978966. Binary fraction =? Since 512/523 = 1 -11/523. 11/523≈0.02103 = binary about 0.000001010110...? Then 512/523 binary = 0.111110101001...? It has zeros within top ~6 bits, so popcount in top 6 is 5 (one zero). Over 513 bits, if random after, zeros ~256, popcount ~257. To have at most one zero among 513 bits, q must be one of 514 numbers (all ones or all ones with one zero). Those are extremely close to 2^{513} (within factor ≥3/4) but not every number above threshold; specifically q must be of form 2^{513}-1 - 2^j for j=0..512 (or 2^{512}-1). That's 514 possible values. Our q for n=523 is unlikely and can be shown not of that form perhaps by modular/division. Need not go there if the actual required distinct substrings force more than weight; maybe we can find exact n via a different interpretation.

Wait: Number of length-10 substrings starting with 1 in a cyclic sequence equals number of ones only if we count starting positions, but distinct required words are all possible 10-bit words beginning with 1. To contain all 512 distinct such words, there must be at least 512 starting positions with bit 1 (yes weight≥512). But if the period block for n≥r+1 has length r and q≈2^r/n. For minimal n maybe r is much larger than 512, and q could have weight≥512 more easily. We want smallest n, not smallest r. Since n≥r+1, to minimize n we want minimal r that can possibly work. Lower bound r≥512. But for r=512, n≥513 and ord_n=512. Then q=(2^{512}-1)/n. Since n≥513, q<2^{512}/513 <2^{503}; bit length≤503 <512, so weight≤503. Impossible. For r slightly larger, q bit length ≈ r - log2 n. Since n≥r+1, bit length ≤ r - floor(log2(r+1)) + maybe. Need this bit length at least 512 (necessary for weight≥512). Solve r - log2(r+1) ≥512. For r=522, r-log2(523)=522-9.031=512.969 -> bit length can be 513. So minimal r for possible weight≥512 is 522 (since for r=521, n≥522? but n odd and ord≤n-1 -> n≥523? Wait if r=521, ord=r implies n≥r+1=522, but n odd ≥523. Then q<2^521/523 ≈2^{511.969}; bit length≤511? Let's compute: 2^521/523 =2^{521}/(2^9.03)=2^{511.97}; floor log =511, bit length=512. Actually bit length can be 512. Then weight≤512; to have weight≥512, q must equal 2^{512}-1 (all lower 512 bits ones) exactly if bit length=512. Is that possible with n=(2^r-1)/q = (2^{521}-1)/(2^{512}-1)? Not integer likely (since 521 not multiple of 512? (2^a-1)|(2^b-1) iff a|b; 512 ∤521, so no). For r=521, q<2^{512}; max integer q≤2^{512}-? Let's exact: q=(2^{521}-1)/n with n≥523? q <2^{521}/523. Compare to 2^{512}: ratio =2^{521}/523 /2^{512}=2^9/523=512/523<1, so q< (512/523)2^{512}=2^{512} - (11/523)2^{512}. This is less than 2^{512}-1? Need check: (11/523)2^{512} ≈0.021032^{512}=2^{506.56} >>1, so q <2^{512} - huge. Thus q has bit length 512 but is significantly below all-ones; popcount≤511? Not necessarily: a number below 2^{512}-huge could still have popcount 512? The only 512-bit number with popcount512 is 2^{512}-1. Since q<2^{512}-1, popcount≤511. Good. For r=521 impossible. For r=522, q<2^{522}/523 = (512/523)2^{513}=2^{513} - (11/523)2^{513}. This is less than 2^{513}-1 by ~2^{507}, so q cannot be all ones; but popcount≥512 would require q be 2^{513}-1 (popcount513) or 2^{513}-1-2^j (popcount512) for some j, or if bit length 512 q=2^{512}-1. Since q is below 2^{513}-1 by D≈(11/523)2^{513}≈? D not a power of two plus/minus. Need prove popcount(q)≤511 maybe by showing q < 32^{511}-1? Wait threshold for at most one zero among 513 bits was q ≥ 32^{511}-1 ≈0.752^513. Our q=0.9792^513 > threshold, so that doesn't rule out. Example q=0.9792^513 could be 2^{513}-1-2^{507}? If subtract a power of two 2^j, ratio deficit=2^{j-513}. Our deficit ratio δ=11/523≈0.02103. For q to have popcount≥512 (bit length 513), deficit D=2^{513}-1-q must be either 0 or a power of two (if exactly one zero). δ≈0.02103 is between 2^-6=0.015625 and 2^-5=0.03125, not a power of two. If D is not 0 or power of two, popcount≤511. We can prove D not power of two via integrality maybe.

But this path is getting complicated and still only necessary (weight) not sufficient for containing all length-10 words. Even if popcount≥512, containing all 512 different 10-bit words starting with 1 in a cyclic sequence of length r with weight≥512. If r=522 and weight maybe 512, then the sequence has exactly 512 ones and each one must start a distinct required length-10 word; since there are 512 required words and 512 ones, every length-10 substring starting at a 1 must be distinct and cover all. That imposes the set of length-10 windows at one positions is exactly all 10-bit strings with first bit1. Is that possible with period 522 and 512 ones? It would mean there are only 10 zeros in the cyclic block (r-weight=10) maybe. A binary cyclic sequence of length 522 with only 10 zeros: can the length-10 windows starting at ones be all distinct? There are 512 ones. But if there are only 10 zeros, among the 512 windows of length10 starting at a 1, by pigeonhole many windows contain no zero? Let's see. A length-10 window starting at 1 may include some of the 10 zeros. Since zeros are scarce, there are long runs of ones. If there is a run of ones of length ≥10, then any window starting at the first through (run_length-9) positions are all "1111111111" repeated, causing duplicates. To have all 512 windows distinct, the word 1111111111 can occur at most once starting at a 1. But with only 10 zeros in length 522, by circular pigeonhole there is a run of ones of length at least ceil((522-10)/10)? Actually zeros separate runs of ones. With z=10 zeros, there are at most 10 runs of ones; total ones=512; average run length 51.2. A run of ones length L contributes max(0,L-9) windows equal to all ones (starting at positions within the run with next 9 bits all one). For L≥10, L-9 duplicates of all-ones. To have all-ones at most once, need every run of ones length L≤10? But total ones≤10*10=100 if z=10 and each run≤10, contradiction. Thus weight cannot be as high as 512 with only r≈522? Wait if weight=512, zeros=10, indeed long runs of ones produce many duplicate all-ones length-10 windows. But required includes word 1111111111 (number 1023) only once; duplicates okay. Duplicates of all-ones don't necessarily prevent containing all other words, but they consume starting positions. We have exactly 512 one-start windows and need 512 distinct words; no duplicates allowed among one-start windows. Because required set size = number of one-start windows (if weight=512). If weight>512, duplicates allowed but still need cover all. For r=522, weight≤513 (bit length upper). zeros≥9. If weight=513, zeros=9, runs of ones total 513 across ≤9 runs average57 -> many duplicate all-ones. Not cover all. If weight=512, zeros=10 same. So r=522 impossible because too many ones relative to zeros creates repeated all-ones? Let's formalize: To have all 512 distinct length-10 words starting with 1, in particular the word 1^10 (ten ones) can occur only once as a start (we need at least once; duplicates not fatal if weight>512, but if weight close to 512 maybe fatal). More generally, runs of ones constrain number of distinct words.

Maybe the minimal construction is different. Let's step back.

Alternate interpretation: The binary decimal representation of 1/n should contain the binary representations of the numbers 1,2,...,1990 as contiguous subsequences, not necessarily all length-10 strings? We reduced all ≤1023 to length-10 strings by extension. That is valid: If the expansion contains the binary representation of every number from 512 to 1023 (all length-10 strings starting with 1), then for any x<512, choose y = x shifted left to length10 (append zeros). y is between 512 and 1023 and its binary representation starts with x's representation; occurrence of y contains x. Good. So requiring numbers 1..1990 is equivalent to requiring: (A) all 10-bit words with leading 1; (B) all 11-bit words with leading 1 and value ≤1990. Note (B) includes some words whose first 10 bits maybe any required length-10 word plus next bit. Since (A) requires all length-10 prefixes u (512..1023). For (B), not all extensions u0/u1 are required, only those with value ≤1990. Let's characterize length-11 required words. They are binary strings of length 11 from 10000000000 (1024) to 11111000110 (1990). In terms of prefix of length? The threshold T=1990 = binary 11111000110. The set of length-11 words ≤T and starting with1. Since all start with1 and length fixed, lexicographic = numeric. It includes all length-11 words w such that w ≤ 11111000110. This is almost all length-11 words with first bits: If w starts with 0? no length11 starts with1. Compare to T: T bits positions 10..0: 1 1 1 1 1 0 0 0 1 1 0. Any w with first five bits <11111 (i.e. among bits10..6 not all ones) is ≤T regardless of rest. Those are numbers 1024..1983? Wait if first five bits (bits10..6) less than 11111, max value with prefix 11110 then all ones lower = 11110111111 =? bits: 1024+512+256+128+0*64+32+16+8+4+2+1 =1983? 1024+512+256+128=1920; +0+32+16+8+4+2+1=1983. Yes all w≤1983 included if prefix not 11111. If first five bits =11111, then remaining six bits must be ≤000110 (since T after five ones is 000110? Let's split T: bits10..6 =11111; bits5..0=000110 (because T=11111000110, after five 1s bits: 0 0 0 1 1 0 = 000110). Values with prefix11111 and suffix ≤000110 (6-bit value ≤6) are included: 1984 + suffix for suffix 0..6? Actually prefix11111 contributes 1984 (since 11111000000=1984). suffix values 0..6 included, suffix=7 (000111) gives 1991 >1990 not included. So required length-11 words = all words whose first five bits are not all ones (1024..1983) plus prefix 11111 with 6-bit suffix 000000..000110 (values 1984..1990). Missing length-11 required? values1991..2047: prefix11111 and suffix 000111..111111 (7..63). Count 57. Good.

Now, if a cyclic sequence contains all length-10 words starting with1, does it automatically contain all length-11 words with first five bits not all ones? Not necessarily; need the correct 11th bit after each length-10 prefix. But maybe requiring length-10 all words and length-11 up to T can be encoded as a shortest common superstring (cyclic) with constraints. Minimal period of a binary cyclic sequence containing a set of words is the length of a shortest cyclic superstring. We need then realize it as 1/n. But the problem asks smallest n, and rational expansions have special form. Maybe there's a known theorem: For any finite set of binary words, there exists n such that 1/n contains them iff no word is all zeros? The minimal n might be obtained by taking a de Bruijn sequence B(k) and setting n=(2^m-1)/A? Hmm.

Let's parse problem statement Chinese: "求最小的正整数n，使得 1/n 的二进制小数表示中能够包含 1, 2, 3, ... , 1990 各个数字的二进制表示。" Maybe from olympiad? It asks smallest positive integer n such that the binary decimal representation of 1/n can contain the binary representations of each of the numbers 1..1990. This sounds like a math contest problem with a neat answer. "能够包含" maybe means the sequence of digits after the binary point contains, as consecutive digits, the representations of all these numbers. There may be a simple answer: n=2^{1990}-1? No.

Maybe they mean the binary decimal representation of 1/n, when written in full (including integer part?), contains the numbers 1,2,...,1990 as in the period? Let's search memory: There is a known problem (IMO shortlist?) "Find the least positive integer n such that the binary expansion of 1/n contains the binary representations of all integers 1,2,...,1990." I vaguely recall an IMO 1990 problem? It might be from IMO 1990 (held in China) Problem? Answer maybe n=2053? Let's recall. IMO 1990 problems: Problem 1 about chords? Problem 2? Problem 3? There is an IMO 1990 Problem 3: Determine all integers n>1 such that (2^n+1)/n^2 integer. That's IMO 1990? Actually IMO 1990 Problem 3: Determine all integers n > 1 such that (2^n + 1)/n^2 is an integer. Yes. Not this.

Maybe from China MO 1990? There is known problem: Find smallest positive integer n such that in binary representation of 1/n, the representations of 1,2,...,1990 all appear. I have a faint memory answer might be 2053? Let's investigate by reasoning, no internet. 1990 suggests use 2^11=2048. The numbers 1..1990 include all binary strings of length ≤10 and most length11. Perhaps minimal n is 2051 or 2053 (prime near 2048) with period something? Wait earlier lower bound using length-10 distinct words gave period r≥512 and n≥r+1≥513. But maybe I erred: A periodic sequence with period r has r starting positions for substrings of a fixed length, but the binary expansion of 1/n may have a non-repeating prefix plus repeating period. We set n odd minimal. If n even, factor 2 shift only, no help. If n has odd part m, the period length r=ord_m(2) can be much larger than m? No r≤λ(m)≤φ(m)≤m-1 for odd m>1. But n=2^a m with m odd; n can be larger than m. Minimal n for a given m is m (a=0). Since r≤m-1, to have r≥512 need m≥513. But wait we only need length-11 representations of numbers up to 1990; maybe the required length-10 strings are not all distinct as substrings because occurrences can overlap with the integer part "1."? For n=1, 1/n=1.000, contains binary representation of 1 as integer part? For n>1, 1/n=0.something. The phrase "1/n 的二进制小数表示" includes the integer part 0? For n=2, 0.1000 contains "1" at first fractional bit. It contains representation of 1, but not 2 ("10")? It has "10" at start yes; not 3. Not relevant.

Could the expansion be non-terminating with preperiod length a and period r; number of possible starting positions for length-10 substrings is infinite, but distinct substrings in an eventually periodic sequence are finite: at most a+r? Actually if preperiod length a and period r, the number of distinct length-k substrings is at most a+r (for k maybe more, but finite). Required 512 distinct length-10 words. If n even with a large power of 2, preperiod consists only zeros? Wait 1/(2^a m) = shift of 1/m, so preperiod before the periodic part may not be all zeros if we choose the canonical eventually periodic expansion? Let's compute exactly: If m odd >1, 1/m=0.\overline{B} purely periodic (no preperiod) because gcd(10? base2, m)=1. Multiplying by 2^{-a} shifts the point right: 0.[a zeros]\overline{B}? Example 1/3=0.010101...; 1/6=0.0010101... yes one zero then repeat 01? Actually 1/6=0.001010101..., preperiod "0" then period "01"? It can be seen as preperiod length1 and period2. The preperiod zeros add only substrings starting with zero or including zeros; required words start with1, so no help. Thus a doesn't increase distinct required words. Good.

But the lower bound r≥512: For an eventually periodic sequence with preperiod length a and period r, the number of distinct length-k substrings that start with 1 could be up to a+r? If the preperiod has ones, but for 1/n with n=2^a m and m odd, preperiod bits are zeros (if we choose period of 1/m shifted). However the canonical binary expansion of a rational with denominator n=2^a m can have preperiod length a and period r=ord_m(2); the preperiod is the first a bits of floor(2^a/n?)? Let's derive: n=2^a m, gcd(m,2)=1. Write 1/n = A/2^a + B/(2^a(2^r-1))? The preperiod bits are not necessarily all zeros? Example n=6 (a=1,m=3): 1/6=0.0010101..., preperiod length a=1 bit is 0, period "01" (r=2) shifted? 1/m=1/3=0.010101; shift right by1 =0.0010101. Yes preperiod zeros. Example n=12=4*3:1/12=0.00010101, two zeros then 01. Good. For m odd, 1/m<1 and purely periodic; shifting right yields a zeros before same period (unless the periodic block begins with zeros, then the distinction between preperiod and period may merge; but canonical minimal preperiod still maybe a? Example m=5, 1/5=0.00110011 (period 0011 starts with zeros); 1/10=0.000110011 (one extra zero). Preperiod bits all zero). Required words start with1; an occurrence cannot start in initial zeros. Could an occurrence start before the periodic part but include the first periodic 1? It would start with 0, invalid because binary representations have no leading zeros. Unless the word is for number? all start with1. So preperiod irrelevant. Good.

But maybe we can choose n not coprime to 2 and with odd part m small but preperiod length a huge containing arbitrary patterns? No, preperiod for 1/n is determined by shifting, always zeros. If n has odd part m=1 (n power of two), expansion terminates: 1/2^a =0.0...01000..., contains only a single 1; no.

So r≥512 stands. But maybe r=ord_m(2) can be ≥512 while m (odd part) is as small as, say, 21? Is ord_m(2)≤λ(m)≤φ(m)≤m-1 always for m>2? φ(m)≤m-1 yes equality prime. λ(m)≤φ(m). Thus r≤m-1. So m≥r+1≥513. Good.

Now minimal odd m with ord_m(2)≥512 is not necessarily 523; we should find smallest odd m≥513 such that ord_m(2)≥512. We listed 513,515,517,519,521,523. But need check 513=2719: ord modulo 27 of 2? 2 is primitive mod 27? ord_27(2)=18? modulo19 ord=18. lcm(18,18)=18. 515=5103: ord5=4, ord103? 102=2317; maybe ord_103(2)=51? lcm ≤204. 517=1147: ord11=10, ord47? 46=223, maybe 23? lcm≤230. 519=3173: ord3=2, ord173 divides172=443, ≤172. lcm≤172. 521 prime: p≡1 mod8, ord divides260, and also p-1=520=2^3513; order could be 260 or 520? Since 2 is quadratic residue mod p (p≡±1 mod8; 521≡1), order divides (p-1)/2=260. <512. 523 prime: p≡3 mod8, 2 nonresidue; order divides522=23^229 and not divide261; order has factor2. To be ≥512, must be522 (as divisor list). Need verify ord_523(2)=522. We can compute 2^{174} mod523 and 2^{18}; if not 1. We already computed 2^18=262144 mod523 =? Let's recompute accurately: 523501 =523500+523=261500+523=262023. 262144-262023=121. Yes 2^18≡121 ≠1. For 2^{174}: Since 174= (p-1)/3. Need show not 1. We can compute via repeated squaring manually maybe. But for final answer we may need rigorous proof not too huge. Could use properties: 523 =? 2^9=512 ≡ -11 mod523. Nice! This is key. Since 2^9 ≡ -11 (mod523). Then we can compute powers. 2^18 ≡121. 2^29? Maybe. We need order. We know 2^9 ≡ -11. Then 2^{18}≡121, 2^{36}≡121^2=14641 mod523. 52328=14644, so ≡ -3. Nice! Then 2^{72}≡9, 2^{144}≡81. Then 2^{174}=2^{144}2^{18}2^{12}? Wait 174=144+18+12. 2^{12}=4096 mod523: 5237=3661 rem435 (or -88). So 2^{174}≡81121435 mod523. 81121=9801; mod523: 52318=9414 rem387 (or -136). Multiply by435: 387435=168345; 523321=167883? compute 523322=168406; rem -61; or 523321=167883 rem462. So 2^{174}≡462 ≠1. Good. Also 2^{261}=-1 by Euler criterion since p≡3 mod8; or compute? We can use Legendre. Then order not divide 261; and 2^{174}≠1, 2^{18}≠1 ensure order not divide proper divisors. Let's structure: p-1=522=2·3^2·29. To prove ord=522, enough show 2^{522/2}=2^{261}≠1, 2^{522/3}=2^{174}≠1, 2^{522/29}=2^{18}≠1. We have 2^{261}≡-1 (since 523≡3 mod8 -> 2 is quadratic nonresidue; or we can compute using 2^9=-11 maybe? Let's see compute 2^{261} from 2^{144}≡81, 2^{72}=9, 2^{36}=-3, 2^{18}=121, 2^9=-11, sum 261=144+72+36+9? 144+72+36+9=261. Product 819(-3)(-11). 819=729≡206; (-3)(-11)=33; 20633=6798; 523*13=6799 -> ≡ -1. Nice! Good, no need Legendre. Great. So ord_523(2)=522. We can prove with 2^9≡-11 and computations. Nice.

But as noted, ord=522 and n=523 may not satisfy substring property due weight/run constraints. Need determine if it does. Maybe the special relation 2^9≡-11 mod523 makes the repeating block nice. Let's explore. For n=523, since 2^9 ≡ -11, we have 523*? =2^9+11=523. Ah! 2^9+11=523. Nice. This relation could make 1/523 have a simple binary expansion! Since 523 = 512+11. Then

1/523 = 1/(512+11) = (1/512) * 1/(1+11/512). Not directly periodic with period? Maybe use 2^9 ≡ -11 => 2^9+11 ≡0. Then 11/523 =? We can derive a repeating block with period 9? No ord not 9. But maybe 1/523 relates to 11/(2^9+11). There may be a series: 1/(2^9+11)=2^{-9} * 1/(1+11·2^{-9}) =2^{-9} Σ_{k≥0} (-11)^k 2^{-9k}. Not a finite binary pattern because negative terms. Alternatively since ord=522=2261 and 2^{261}≡-1, we have 2^{522}-1=(2^{261}-1)(2^{261}+1), and n divides 2^{261}+1 (because 2^{261}≡-1). Then 1/n = A/(2^{261}+1) maybe expansion with period 522: bits = A/(2^{261}+1) = A2^{-261}/(1+2^{-261}) = A(2^{-261} -2^{-522}+2^{-783}-...) not simple. But 2^{261}≡-1 mod523 means 523 divides 2^{261}+1. Quotient B=(2^{261}+1)/523. Since 2^{261} huge, B≈2^{261}/523≈2^{251.97}, bit length 252. Then the 522-bit repetend q=(2^{522}-1)/523 = B(2^{261}-1). In binary, B (252 bits) times (2^{261}-1) (261 ones) yields q = B shifted left 261 minus B. This has a structured form: q = (B <<261) - B = (B-1) followed by? Let's derive: B*(2^{261}-1)= (B-1)2^{261} + (2^{261}-B). If 1≤B≤2^{261}, then as a 522-bit block (since B bit length 252, B-1 bit length ≤251, and complement 2^{261}-B is 261 bits), the repetend consists of (B-1) in high 251? Let's compute: (B-1)2^{261} + (2^{261}-B). The second term C=2^{261}-B is between 1 and 2^{261}-1, represented in 261 bits; since B≈2^{252}, C≈2^{261}-2^{252}, which in 261 bits starts with? C = 2^{261}-B = (2^{261}-1) - (B-1). If B-1 has bit length ≤252, then C as 261-bit string has leading (261-252)=9 ones then complement of B-1 within lower252 bits? Wait B≈2^{251.97}? Let's compute B=(2^{261}+1)/523 ≈2^{261}/523 =2^{261}/(2^9+11). Ratio to 2^{252}: 2^{261}/523 /2^{252}=2^9/523=512/523=0.978966. So B≈0.9789662^{252}; bit length 252, top bits same fraction 111110... Then B-1 <2^{252}. The high part (B-1) occupies bits positions 261..511 (251? Actually q has 522 bits positions 0..521. (B-1)2^{261}: B-1 up to <2^{252}, shifted by261 gives bits up to position 512 (since 252+261-1=512), leaving top positions 513..521 zero (9 zeros). Good: repetend has 9 leading zeros (matches first one at bit10). Low part C=2^{261}-B is a 261-bit number. Since B<2^{252}, C = 2^{261}-B has top bits positions 260..252 all ones? Because subtract B (bit length252) from 2^{261}: 2^{261}-B = 2^{252}(2^9) - B = 5122^{252} - B. If B≈0.9792^{252}, then C≈(512-0.979)2^{252}=511.0212^{252}, bit length 261 (since >5112^252 >2^260). The top 9 bits of C maybe all ones except slight. More exact: C = (2^{261}-1) - (B-1). B-1 is a 252-bit number. Complement within 261 bits: leading 9 bits (positions260..252) are ones (because B-1 has zeros there), and lower252 bits are bitwise complement of B-1. Thus the 522-bit repetend is: [9 zeros][B-1 (252 bits)] [9 ones][complement of B-1 (252 bits)]? Let's check lengths: high part B-1 shifted by261: if B-1 has bit length 252, it occupies bits 512 down to261 (252 bits). Before it bits521..513 are 9 zeros. Low part C occupies bits260..0 (261 bits). C = complement of B-1 in 252 bits plus leading 9 ones at bits260..252. So total repetend length 9+252+9+252=522. Nice! The block is 0^9 (B-1) 1^9 (~(B-1)). Where B-1 is 252 bits. This structure might help determine occurrences. B-1≈0.9792^252 -1. Its binary begins with? B=(2^{261}+1)/523. Since 523=512+11, maybe B has a simple form. Let's compute B exactly maybe via division using 2^9≡-11. 2^{261}=2^{929}=(2^9)^{29}≡(-11)^{29} mod523. Since 2^{261}≡-1, (-11)^{29}≡-1 -> 11^{29}≡1? Actually (-11)^29 = -11^29 ≡ -1 -> 11^29≡1 mod523. Not simpler.

Maybe B = (2^{261}+1)/523. Because 523=2^9+11, we can perform polynomial division: Let x=2^9=512. Denominator x+11. Numerator x^{29}+1 (since 261=9*29). x^{29}+1 divided by x+11. Since 29 odd, x^{29}+1=(x+11)(x^{28}-11x^{27}+11^2 x^{26}-...-11^{28})? For odd 29, x^{29}+11^{29} divisible by x+11, but numerator is x^{29}+1, not x^{29}+11^{29}. Mod x+11, x≡-11, x^{29}+1 ≡ (-11)^{29}+1 = -11^{29}+1. For this to be 0 mod integer x+11=523, need 11^{29}≡1 mod523 (true if ord?). Not exact polynomial with integer quotient because constant remainder? But since 11^{29}≡1 mod523, x^{29}+1 = (x+11)Q + R where R=1-(-11)^{29}=1+11^{29}? Wait substituting x=-11 gives (-11)^29+1 = -11^29+1. If this is divisible by523, yes. Quotient Q coefficients alternate with powers 11 but final remainder adjusted. Not a simple 0/1 pattern.

This is getting deep. Maybe the answer is not 523. Let's search memory mentally. I recall a known problem with answer n=2^{11}-1=2047? Because to contain binary representations of 1..1990, it suffices that the repeating block is 11111111111 (11 ones)? No. Maybe n=2047 gives 1/2047 = 1/(2^{11}-1)=0.\overline{00000000001} (period 11, a single 1). That contains only sparse ones; not all.

What if choose n = 2^{2048}-1? The expansion is a single 1 every huge period; not contain.

Maybe the problem means something else: "1/n 的二进制小数表示中能够包含 1,2,...,1990 各个数字的二进制表示" Could mean in the binary representation of the fraction 1/n, the digits after the point include the sequence of binary representations of 1,2,...,1990 consecutively? Like the concatenation 1 10 11 100 ...? No, "各个数字" maybe "the digits 1,2,3,...,1990" but in binary there are only digits 0 and1. In Chinese, "各个数字的二进制表示" = the binary representations of the numbers 1,2,...,1990. Yes substrings.

Maybe the intended trick: It is enough and necessary that the period (repetend) have length at least 1990? Wait to contain the binary representation of k for each k≤1990, because these representations include all strings of length up to 10 and some length 11. A theorem by? The repetend of 1/n is the binary expansion of (2^r-1)/n. If we take n=2^{11}-1=2047, repetend length 11 with a single 1. Not.

Could there be a small n whose non-periodic part (due to denominator factors other than 2?) In base 2, rational denominator n may have factors; if n has odd part m, expansion eventually periodic. Only factor 2 gives preperiod zeros. No other preperiod patterns. So small n cannot have long period.

Let's compute lower bound more carefully maybe using the largest number 1990 (length 11) and all numbers from 1024 to1990? There are 967 required length-11 words. A period-r sequence has at most r distinct length-11 substrings. Therefore r≥967! Ah! I previously used length-10 count 512, but required length-11 words count =1990-1024+1=967. That's much stronger. Since all these length-11 binary representations are distinct and must appear as length-11 substrings. A cyclic period of length r has at most r distinct length-11 substrings (start positions). The preperiod zeros don't help. Hence r≥967. Then n≥r+1≥968; n odd => n≥969. This changes everything. Minimal n at least 969. Need check odd n from 969 upward with ord_n(2)≥967. Since ord_n(2)≤λ(n)≤φ(n)≤n-1. For prime n, need n-1≥967 -> n≥968; smallest odd prime ≥969 is 971? 969=31719? composite; 970 even; 971 prime? sqrt31.1; test divisibility: 971 mod3=17? 9+7+1=17 no; mod5 no; 7:7138=966 rem5;1188=968 rem3;1374=962 rem9;1757=969 rem2;1951=969 rem2;2342=966 rem5;2933=957 rem14;3131=961 rem10. prime. For p=971, p-1=970=2597. Need ord_971(2)≥967. Divisors of970:1,2,5,10,97,194,485,970. The only divisor ≥967 is970. So need ord=970 (2 primitive mod971). Is it? Need check 2^{970/q}≠1 for q=2,5,97: 2^{485},2^{194},2^{10}. 2^{10}=1024≡53 mod971 ≠1. Need 2^{194} and 2^{485}. p=971 mod8 = 3 (968 divisible by8, rem3), so 2 is nonresidue -> 2^{485}≡-1 ≠1. Good. Need 2^{194}≠1 (q=5? Wait 970/5=194) and 2^{10}≠1 (970/97=10). We need compute 2^{194} mod971. Could use relation? 971=1024-53 =2^{10}-53. Not as neat. Maybe there is a closer candidate with nice relation like 2^{10}+?=n. We need smallest n; check composites 969, 970 even, 971. If 971 works for order and substring property maybe answer 971? But does repetend contain all required length-11 words? Period r=970 equals exactly number of required length-11 words (967) plus 3 extra. Very tight. It seems unlikely but maybe the repetend of 1/971 has a special property. Let's examine count: Required length-11 words count = 967 (1024..1990). A period of length r=970 has 970 length-11 substrings. To contain 967 distinct required words, almost all length-11 substrings must be required; only 3 can be nonrequired or duplicates. But required length-11 set excludes all words with prefix11111 and suffix>6 (57 missing) and includes many. For the sequence to include 967 distinct required length-11 words among 970 windows, it must avoid almost all missing words. In particular, among the 57 missing length-11 words (all starting 11111000111 through 11111111111), at most 3 distinct can appear (unless duplicates). This imposes that windows starting with prefix 11111 are heavily restricted. Could the repetend of 1/971 achieve this? Maybe not. The lower bound r≥967 may not be sufficient; the exact minimal n might be much larger or have a nice form. Let's analyze structurally.

Observation: Required set includes all length-11 words except those with first five bits 11111 and last six bits ≥000111 (i.e. suffix 7..63). Missing set M = {11111 s : s is 6-bit, 7≤s≤63}. Equivalently all length-11 words that begin with 11111000111? Let's find common prefixes of missing words. Missing values 1991..2047. Binary 1991 =11111000111; 2047=11111111111. These are all length-11 words with prefix 11111 and not suffix 000000..000110. That set can be described as words containing? The smallest missing is 11111000111. It has first eight bits 11111000? T=11111000110. Missing are T+1..2047. They all start with 1111100? Let's see T prefix first 8 bits =11111000 (since T=11111 000110; first eight: 11111000). Missing values from1991 to2047 include 11111000111 (first8=11111000), then 11111001000 (first8=11111001), ... up to11111111111. So missing set includes some with first8=11111000 (only suffix low3=111? Actually first8=11111000 corresponds bits: prefix11111 + bits5..3=000, low bits2..0. T has low3=110 (6). Missing with same first8 and low3=111 only: 11111000111 (1991). Then first8=11111001 (values1992-1995? low2 00..11) etc. Not a single prefix.

Maybe containing all required length-11 words is equivalent to containing all binary strings of length 11 that do NOT have a certain property. But by extending shorter words, maybe requiring numbers up to1990 actually forces all length-11 words except the all-ones? Let's check: Required numbers include 1984..1990 (11111000000 to11111000110). If a sequence contains all length-10 strings starting with1 and also the particular length-11 word 11111000110 (1990), does that plus containing length-10 strings force missing length-11 words? No.

Maybe use occurrences of smaller numbers to cover missing length-11? Not required. We just need minimal n; maybe we can construct n via repetend equal to a de Bruijn sequence of order 11 (cyclic) with all 2048 length-11 words. If the repetend is a de Bruijn sequence B(2,11) of length2048, it certainly contains all required. Can we realize a de Bruijn sequence as 1/n? The repetend of 1/n has period r=ord_n(2). If r=2048 and repetend block is de Bruijn of order11, then n=(2^{2048}-1)/q where q is the integer whose 2048-bit binary representation is the de Bruijn block. For n to be minimal and odd, need q=(2^{2048}-1)/n integer and ord_n=2048. The smallest n with period 2048 is at least2049 (since ord≤n-1), odd ≥2051. Is there n=2051 with ord2048? 2051? Factor? 2051=7293? 7293=2051. λ=lcm(6,292)=292 <2048. 2053 maybe prime? 2053 sqrt45.3; test primes: mod3 2+0+5+3=10 no;5 no;7 7293=2051 rem2;11186=2046 rem7;13158=2054 rem-1;17120=2040 rem13;19108=2052 rem1;2389=2047 rem6;2970=2030 rem23;3166=2046 rem7;3755=2035 rem18;4150=2050 rem3;4347=2021 rem32. prime? If p=2053, p-1=2052=2^23^319? 2052/4=513=2719. Max divisor ≥2048? Only 2052? But ord divides2052; to be ≥2048 must be2052. But p=2053 ≡5 mod8? 2053 mod8=5, so 2 is nonresidue? For p≡3 or5 mod8, (2/p)=-1. Then 2^{1026}≡-1, order has factor4? p-1 has factor4 yes. Need check other divisors; maybe 2 primitive? Not sure. But a period r=2052 for p=2053 has only 2052 length-11 windows, enough for 967 required and also all length-10 via extension. More plausible. But why 1990 then? If minimal answer were a prime just above 2048, the use of 1990 (not 2047) would matter only to reduce required length-11 count to 967, leading lower bound n≥971 maybe, not 2053. Need identify intended answer.

Let's search memory more: There is a classic problem from 1990 IMO (maybe shortlist) with answer n=2^{11}+1=2049? Hmm. "Find the smallest positive integer n such that the binary representation of 1/n contains the binary representations of all integers from 1 to 1990." I faintly remember an answer involving n=2049? Let's test: n=2049=3683. ord? mod3 ord2; mod683? 682=21131. ord_683(2)? Since 683 =? 2^? 2^10=1024≡341; 2^11≡682≡-1 mod683! Ah! 683 divides 2^{11}+1=2049. Indeed 2049=3683, and 2^{11}≡-1 mod683? Check 2048+1=2049 divisible by683? 683*3=2049. Yes modulo683, 2^{11}=2048≡-1. Then ord_683(2)=22 (if not smaller). lcm with mod3 ord2 =22. So period r=22. The repetend length 22 cannot contain 967 distinct length-11 words (only22 windows). So not.

Maybe the problem's phrase "包含 1,2,...,1990 各个数字的二进制表示" could mean the fraction's binary digits contain the decimal digits 1,2,...,1990? No.

Let's infer intended solution from the number 1990. Since 2^{10}=1024 <1990<2048=2^{11}. Binary representations of 1..1990 include all strings of length≤10 and length11 up to11111000110. A key observation: The binary representation of any integer m with 1≤m≤1990 appears as a substring of the binary representation of some integer among 1024..1990? Let's see for m<1024, extend to length11 maybe by appending bits to get a number between1024 and1990. If m has length l≤10, append (11-l) bits. Can we always choose appended bits so the resulting 11-bit number ≤1990? Since any 11-bit number whose first bit after? If m small, appending zeros yields x2^{11-l}; for m=1 (binary1), append ten zeros ->1024 (included). For m up to1023, shifting left to length11 yields between1024 and2046; some exceed1990 (if m large). But we can choose appended bits not all zeros to stay≤1990? If m has length10 (512..1023), shifting left one yields1024..2046. For m>995? x2 may exceed1990. But we can choose the length-11 number with prefix maybe not start with m? To contain m as substring, we can find a required length-11 word w (≤1990) that contains m's binary representation. Is that always possible? For m length10 starting with1, if x≤995 then x0 (append0) ≤1990; if x≥996, x0≥1992 not included, x1 bigger. Could m appear not as prefix but inside an included length-11 word? Example x=1023 (1111111111) is length10 all ones. Does any required length-11 word ≤1990 contain ten consecutive ones? Required words ≤11111000110 have at most five consecutive leading ones then zeros; none contain 1111111111. But number1023 itself (binary1111111111) is required and length10; if not contained in any length11 required word, it must appear by itself in the expansion. It can appear across a boundary: a length-10 occurrence need not be extendable to a required length-11 word. So our reduction to length11 required words as containers for shorter words may fail for large m. But for counting distinct length-11 substrings, requiring numbers1024..1990 (967 distinct) is valid regardless. The period must have at least967 distinct length-11 substrings, so r≥967. Good. Additionally, number1023 (ten ones) must appear; it may appear as substring crossing between length-11 windows; not an issue.

Can r=967 possible? Then n≥968 (odd≥969). If n=971 prime with r=970, the period has exactly970 length-11 windows. Need contain 967 distinct required words. Is that feasible for a rational repetend? Maybe the repetend of 1/971 (if primitive) could be a de Bruijn-like sequence of order11 missing only 3 windows, but the missing required? Let's count missing length-11 words overall: There are 2^{10}=1024 length-11 words starting with1. Required 967, missing57. A period970 sequence has at most970 distinct length-11 words; to include 967 required, it can include at most3 extra distinct words. Therefore among the 57 missing words, at most3 distinct can occur. In particular, the word 11111111111 (2047) cannot occur unless it's one of ≤3 extras; but even if it occurs, it would imply occurrence of ten consecutive ones (1023) and eleven ones. Is that bad? Not directly. But the set of length-11 windows of a binary cycle of length970 that avoids 54 specific words and includes 967 others: Since there are only 970 windows, and need 967 distinct required, the sequence's length-11 windows must consist of 967 required distinct words plus at most3 windows that are either duplicates or missing. That means almost all length-11 windows are distinct and required. In particular, the graph of overlapping length-11 windows (a walk in de Bruijn graph dimension10) of length970 visits 967 distinct edges (required edges) in the de Bruijn graph (vertices =10-bit strings; edges=11-bit strings). Required edges are those from a 10-bit vertex u to v (shift left + bit) with edge label ≤T. Missing edges are 57 specific edges all originating from vertices with prefix11111? Let's understand de Bruijn graph: vertices are 10-bit strings. Required length-11 words starting with1 are edges whose first bit=1 and label≤T. Since first bit=1, the source vertex u (first10 bits) ranges over all 10-bit strings starting with1 (512 vertices). For each source u, there are two edges u0,u1. Required includes both except when the resulting 11-bit word >T. The missing edges are exactly the 57 edges with labels1991..2047. They all start with bits11111; their source vertices are the first10 bits of these labels: for labels 1991..2047, source = floor(label/2) from995 to1023. Specifically missing edges are from source vertices u=995..1023 (binary 1111100011? to1111111111) with certain outgoing bits. Let's list: label even=2u, odd=2u+1. Missing labels1991..2047. For u=995 (1111100011), labels1990 (required) and1991 missing: so edge u->1 missing, u->0 required (1990). For u=996..1023, both labels 2u and2u+1 are ≥1992 missing (since u≥996 =>2u≥1992). That's sources996..1023 (28 vertices) both outgoing edges missing =56 edges, plus the edge from995 to1 =1, total57. Good. Required edges from source vertices512..1023: For sources512..994 (483 vertices) both outgoing edges required (labels1024..1989). For source995, only edge0 required (label1990), edge1 missing. For sources996..1023 (28 vertices) no outgoing edges required.

Now in the cyclic sequence, length-11 windows correspond to a closed walk of length r in the de Bruijn graph on 10-bit vertices. Required windows = required edges traversed. To include all 967 required edges in a walk of length r≥967. If r=967 exactly, the walk must traverse each required edge exactly once: an Eulerian circuit in the subgraph of required edges. Is the required-edge subgraph Eulerian (balanced indegree=outdegree at each vertex, strongly connected)? If yes, then a period of length967 containing all required edges exists (as a sequence), but can it be 1/n? Let's check degrees. Vertices are all 10-bit strings (including those starting with0?) But required edges have first bit1, so their source vertices start with1. However the target vertex of an edge (last10 bits) may start with0 or1. The closed walk may pass through vertices starting with0 via targets, but the next edge from a vertex starting with0 would have first bit0, which is not a required length-11 word (numbers length11 start with1). Could such an edge appear in the expansion? If r minimal=967 and we need traverse all required edges exactly once, there is no room for edges starting with0. Therefore every edge in the cycle must be required, so every vertex (the length-10 window at each position) must be the source of a required edge and hence start with1. But a binary cycle where every length-10 window starts with1 has all bits1 (if r≥10), impossible. More carefully: If all r length-11 windows are required (start with1), then every starting bit of a length-11 window is 1; there are r starting positions covering all bits of the cycle, so the entire cycle is all ones. Then the only length-11 window is 11111111111, not required. Contradiction. Therefore r cannot equal number of required edges; the sequence must include some non-required windows/edges (with first bit0 or missing) to transition between vertices. Indeed to have a length-11 window starting with1 at position i, bit i=1. For consecutive windows, bit i+1 may be0 or1. Vertices (length-10 windows) can start with0, but then the next edge (length-11 window starting at i+1) starts with0, which is not required; such windows must be extra. Since the cycle cannot be all ones, there must be zeros; each zero bit starts a non-required length-11 window. Let z = number of zeros in period. Then number of length-11 windows starting with1 = r-z = weight w. Required edges count967 must be ≤ w (distinct), so w≥967. Thus z≤r-967. For r=970, z≤3. A binary cycle of length970 with at most3 zeros. But as before, with at most3 zeros there are long runs of ones, causing many repeated length-11 windows equal to all ones. More importantly, the required edges include sources512..994 etc. The subgraph of required edges has outdegree 2 for sources512..994, outdegree1 for source995, outdegree0 for sources996..1023. If every required edge must be traversed, sources996..1023 have outdegree0 in required subgraph but they appear as targets? Vertex1023=1111111111: required edge11111111110? label2046 missing, and11111111111 missing. So any occurrence of vertex1111111111 as a length-10 window means the next bit produces a missing edge. Vertex1111111111 must occur as the last10 bits of some required edge? Which required edges end with target1111111111? An 11-bit required label whose last10 bits are all ones would be b1111111111. If b=0 label=01111111111=1023 (length? as 11-bit with leading0 not a number representation; but length-11 window starting with0 nonrequired), if b=1 label=2047 missing. No required label ending in ten ones. Thus the length-10 window 1111111111 cannot be the target of a required edge. However the required word for number1023 is ten ones; it can occur as a length-10 substring. Its occurrence in the cycle corresponds to some starting position i with bits i..i+9 all ones. Then the length-11 window starting at i is either 11111111110 or11111111111, both missing (or if i is such that previous?); so an extra missing edge is unavoidable. That's okay (at most3 extras for r=970). But vertices996..1023 have outdegree0 required; if the walk enters such a vertex (length-10 window ≥111110001100? actually 996..1023) after traversing a required edge, it must leave via a nonrequired edge. Which required edges target these high vertices? A required edge label L≤1990 has target t = L mod1024 (last10 bits). For L in1024..1990, t ranges0..966. Specifically targets of required edges are L-1024 for L=1024..1990: t=0..966. So no required edge targets a vertex ≥967. Aha! The length-10 windows (vertices) that are sources of missing edges (995..1023) are never targets of required edges. But in a cyclic sequence, if such a vertex occurs as a length-10 window, the preceding length-11 window (ending at that vertex) would have label with last10 bits = vertex. If vertex≥967, that preceding label = b*1024 + vertex. If b=0 label=vertex (0..1023) which as 11-bit starts with0 and value<1024 (not a required length-11 number; but could be nonrequired window starting0). If b=1 label=1024+vertex ≥1991 for vertex≥967 (missing). So any occurrence of a length-10 window v≥967 forces the next edge (starting at v) and the previous edge (ending at v) to be non-required (previous either starts with0 if b=0 or missing if b=1). This may create many extras.

But do vertices ≥967 need to occur? Yes, as sources of required edges? Required edges sources are512..995 (both edges for512..994, edge0 for995). Sources996..1023 have no required outgoing, so they need not occur as starts of required windows. But they might occur as targets of required edges? We found required labels L=1024..1990 have last10 bits t=0..966, so targets only0..966. Therefore no required edge ends at a high vertex ≥967. Hence in a walk consisting only required edges plus minimal connectors, high vertices ≥967 are isolated from required subgraph (no required in/out). They should not occur if we want minimize extras, except the length-10 word1111111111 (1023) is required as number1023. Wait number1023 binary length10 =1111111111. It must appear as a substring. That is exactly vertex1023 occurring starting at some position. As argued, this forces surrounding length-11 windows to be nonrequired. It also implies a run of at least ten consecutive ones. With z small, runs of ones inevitable. But maybe one run of ten ones can serve number1023 and also contain many required length-11 words? A run of ones length L produces length-11 windows all ones for starts within run (missing) and windows with a single zero at the end/beginning near boundaries. Required words mostly have first five bits not all ones, so they need zeros relatively early. If the cycle has weight≥967 and r maybe around? Let's derive lower bounds using zeros and required edges count more stringently.

Let the binary cycle (repetend) have length r and z zeros, w=r-z ones. Required length-11 words count E=967. Each occurrence of a required length-11 word starts at a one. Distinct required words E≤w. Hence z≤r-E. Also, consider runs of ones. If there are z zeros (z>0), there are z runs of ones (circular) some possibly length0 if zeros adjacent. Let run lengths ℓ_1,...,ℓ_z (nonnegative, sum w). A run of ones length ℓ produces max(0, ℓ-10) length-11 windows consisting of eleven ones? More relevant: The required set includes no word with first six bits all ones? Required length-11 words have first five bits not all ones OR prefix11111 with suffix≤000110. In particular, any required word cannot contain six consecutive ones anywhere? Let's check required length-11 words ≤11111000110. The maximum run of consecutive ones in such a word: It starts with at most five ones (11111) followed by 0. Could there be six consecutive ones spanning later bits after some zeros? To have six consecutive ones not at the very start, the word would contain a substring 111111 of length6. Would that force value >T? Not necessarily e.g. 10111111000 (value? first bit1, second0 => <11111... so ≤1983) contains six? bits positions: 1 0 111111 000 has six ones after a zero, value =1024+0+256+128+64+32+16+8=1528? It is ≤1990 and required. So required words can contain runs of ones after an early zero. But any required length-11 word must have a zero within the first? If first five bits all ones then sixth bit must be0 (T prefix111110...; actually required with prefix11111 have suffix000000..000110, so bits6..8 =000). If first five bits not all ones, there is a zero among bits2..5. In all cases, every required length-11 word has at least one zero among its first six bits? If first five bits=11111, sixth bit=0 yes. If not, there is a zero among first five. So yes: every required length-11 word contains a zero in its first six bits. Equivalently, a length-11 window starting at a one can be required only if among the next five bits (positions i+1..i+5) there is a zero; i.e. the run of ones starting at position i has length at most5? Let's verify: If bits i..i+5 are all ones (six consecutive ones starting at i), then the window's first six bits are111111, which is >111110... and definitely missing (≥2016? Actually 11111100000=2016 >1990). Such a window cannot be required. Therefore any starting position i whose following run of ones has length at least6 (bits i..i+5 all ones) yields a non-required length-11 window. To have 967 distinct required windows, we need at least967 starting positions i with bit i=1 and run length from i ≤5 (i.e., a zero occurs within the next five bits). Let call such positions "good". Good one positions are exactly ones that are within distance≤5 after a zero? In a run of ones of length ℓ, the starting positions in that run that are good are those with at most5 consecutive ones before the next zero: if ℓ≤5, all ℓ positions good; if ℓ>5, only the last5 positions of the run (closest to the following zero) are good? Let's test: run bits positions 1..ℓ all ones, next zero at ℓ+1. For start position j in run, the window bits j..j+10. It is required only if among first six bits j..j+5 there is a zero. Since within run all ones until ℓ, need ℓ+1 (zero) ≤ j+5, i.e. j ≥ ℓ-4. So in a run length ℓ, good starts are the last min(ℓ,5) positions? If ℓ≤5, all ℓ starts good (zero within first ℓ+1≤6 bits). If ℓ>5, only starts j=ℓ-4,...,ℓ (5 starts) have zero within first six bits. Earlier starts have six consecutive ones -> window begins111111, not required. Thus each run of ones contributes at most5 good starting positions. Excellent! Therefore the number of required length-11 windows ≤ total good starts ≤5 z (where z = number of zero runs = number of zeros if no adjacent? Actually number of runs of ones ≤ number of zeros z if z>0; each zero can close at most one run; with z zeros there are at most z runs of ones). More precisely if z>0, circular sequence has at most z runs of ones (exactly z if no adjacent zeros? If zeros adjacent, some runs length0; number of positive one-runs ≤z). Each positive run contributes at most5 good starts. Hence number of good starts ≤5z. We need at least967 distinct required length-11 windows, each starting at a good one. Are distinct required windows could start at same position? no. So 967 ≤ #good starts ≤5z. Hence z ≥ ceil(967/5)=ceil193.4=194. Great! Then w=r-z. Also E≤w. For minimal n, r=ord_n(2)≥? We need r = w+z ≥ E + z ≥967+194=1161. Since r≥1161. Then n≥r+1≥1162; n odd≥1163. This is a stronger lower bound. Nice. Maybe intended answer around1163? Let's refine.

We used fact every required length-11 word has a zero among first six bits. Let's confirm for all numbers1024..1990. Binary length11. If the first six bits were111111, value ≥11111100000 = 2016? Compute bits10..5 all ones: contribution 1024+512+256+128+64+32=2016, plus lower bits ≥0 =>≥2016 >1990. Yes. So required words cannot start with six ones. Therefore a start position in a run of ones must be followed by a zero within five bits. In any run of ℓ consecutive ones bounded by zeros (cyclic), at most min(ℓ,5) starting positions yield required words. If ℓ large, only last five starts. Good. Number of one-runs ≤ z (number of zeros). Thus required windows ≤5z. Since there are967 required numbers with 11-bit representations (1024 through1990), z≥⌈967/5⌉=194. Also to have 967 distinct required windows, need ones count w≥967 (trivially if z≥194 and r maybe). Then period r=w+z≥967+194=1161. Actually w could be less than967? No each required window starts at one, distinct starts, so w≥967. So r≥1161.

Then n (odd part) satisfies ord_n(2)=r≤λ(n)≤φ(n)≤n-1 for odd n>1, so n≥r+1≥1162. Since n odd, n≥1163. Candidate smallest n=1163? Need check factorization and order. 1163? sqrt≈34.1. Test primality: mod3:1+1+6+3=11 no; mod5 no;7:7166=1162 rem1;11105=1155 rem8;1389=1157 rem6;1768=1156 rem7;1961=1159 rem4;2350=1150 rem13;2940=1160 rem3;3137=1147 rem16. prime? yes 1163 seems prime. p-1=1162=2783? 1162/2=581=783. Divisors:1,2,7,14,83,166,581,1162. Only divisor ≥1161 is1162. So if n=1163 prime, ord_n(2) must be1162 to reach r≥1161 (since proper divisors≤581). Need check if 2 is primitive mod1163. p mod8: 1163 mod8 = 3 (1160 divisible by8? 8145=1160 rem3), so 2 nonresidue -> 2^{581}≡-1 ≠1 (good for q=2). Need check 2^{166}≠1 (1162/7=166) and 2^{14}≠1 (1162/83=14). 2^{14}=16384 mod1163: 116314=16282 rem102 (or -1061) ≠1. Need compute 2^{166} mod1163. We can do by repeated squaring. Let's compute later. If primitive, r=1162. Then lower bound r≥1161 satisfied with z≥194, w=r-z. For n=1163, q=(2^{1162}-1)/1163 ≈2^{1162}/1163. Since 1163 between1024 and2048 (2^10=1024,2^11=2048), 1/n between2^-11 and2^-10; first 10 bits? n>1024 =>1/n<2^-10, n<2048=>1/n>2^-11, so first ten fractional bits are zero and bit11=1. The repetend q has bit length? q<2^{1162}/1024=2^{1152}; q>2^{1162}/2048=2^{1151}; so bit length1152. The r-bit block has r - bitlength =1162-1152=10 leading zeros. So zeros count at least10 only; but our lower bound requires z≥194 zeros in the repetend. Does q have at least194 zeros? Likely yes (~half of1152 bits ≈576 zeros). Not guaranteed but probable. More importantly, can the repetend contain all required length-11 words? Period r=1162. Required E=967. Lower bound says need z≥194; r-E=195 extra windows. If z=194 and w=968 maybe tight. For n=1163, r=1162, so to satisfy we must have z between194 and195? Since w≥967 => z≤r-967=195. And z≥194. Thus the repetend must have exactly z=194 or195 zeros, and w=968 or967 ones. But q bit length is1152 within a 1162-bit block (10 leading zeros). Having only194/195 zeros total among1162 bits means q (1152 bits) has only184/185 zeros, i.e. q has Hamming weight about967/968 ones out of1152 bits. Is that possible? q≈(2^{1162})/1163. Ratio to 2^{1152}: 2^{10}/1163 =1024/1163≈0.880481. So q≈0.88052^{1152}. A random 1152-bit number around0.882^1152 has expected ones ≈? The leading bits: 0.8805 binary =111000011...? Let's compute 1024/1163 =1 -139/1163 =1 -0.119518. Binary of 0.8805 begins 0.111000010... maybe. It will have roughly half ones, ~576, not 968. To have only185 zeros among1152 bits, q would need be extremely close to all ones (deficit with ≤185 zeros). Not the case. So n=1163 cannot work. Our lower bound r≥1161 is necessary but for n≈r+1≈1163, q has bit length only r-10, and to have z≥194 zeros total, q must contain at least184 zeros among its1152 bits. That's plausible (expected576), not impossible. Wait I reversed: z total≥194; leading zeros=10, so q must have ≥184 zeros. That's easy. I mistakenly said only194 zeros total; that's the minimum, not exact. For n=1163, r-E=195, but z can be much larger than194; if z large, w=r-z might drop below967. Need w≥967 imposes z≤195. Right! Because r fixed1162 and required E=967 distinct windows each start at a one; if zeros z≥196, ones w≤966 <967 impossible. Therefore z must be ≤195. Combined z≥194 -> z is 194 or195. That is extremely restrictive. For q with bit length1152 (10 leading zeros), total zeros z =10 + zeros_in_q (where zeros_in_q among1152 bits) must be ≤195 -> zeros_in_q≤185. So q must have at least967 ones among1152 bits. Is q≈0.88052^1152 capable of having ≥967 ones? That would mean among1152 bits only≤185 zeros. But q is less than0.88052^1152; numbers with ≤185 zeros can be as low as? The smallest 1152-bit number with at most185 zeros (i.e. ≥967 ones) is obtained by placing zeros in the highest possible positions to minimize value? To minimize value with bit length1152 (top bit1) and at most185 zeros among lower1151 bits, put all185 zeros immediately after the leading1 (bits1150 down to966) and ones thereafter. Value ≈2^{1151} + (2^{966}-1) ≈2^{1151}+2^{966}, which relative to2^{1152} is0.5 + tiny. So the count condition alone does NOT require q close to all ones; q=0.88052^1152 could have ≥967 ones if its zeros are concentrated in high positions. But q's binary leading fraction1024/1163≈0.8805 = binary? Let's compute: 0.8805 decimal is 111000011... It has three zeros in the first six bits, not zeros concentrated then all ones. After the leading bits, likely random ~half zeros. Expected zeros ~ (1152 - maybe leading pattern)/2 ≈570, far >185. We need a rigorous way to show q has many zeros, or more directly show period r=1162 cannot contain all required words unless zeros≤195, and then show for n near1162 the repetend has >195 zeros. Perhaps use q < (1024/1163)2^{1152}; the first few bits of q are fixed and include zeros, but only a few. Need lower bound on zeros of q. Since q=(2^r-1)/n. Maybe using modulo small numbers, the repetend cannot have too few zeros. There may be a general argument: If a binary cycle of length r contains all required length-11 words, then z≥194 and w≥E. For r minimal 1161 or1162, w=E or E+1, z=194/195, so the sequence is extremely constrained: almost every one starts a required edge and there are exactly about194 zeros. Could such a sequence exist at all? Let's analyze combinatorially independent of rational. Maybe minimal period for a binary cycle to contain all required length-11 words is much larger than1161 due degree balance/Eulerian connectors. Let's model as de Bruijn graph and find shortest closed walk covering all required edges (rural postman). Then among sequences with that period, find smallest n whose repetend realizes it? Maybe the rational constraint can be satisfied by choosing n equal to a divisor associated with the constructed superstring? But problem asks smallest n; maybe by a theorem the minimal n corresponds to the shortest possible period, and the repetend can be chosen via n=(2^r-1)/A? Need investigate.

Let's focus on the combinatorial problem: Find minimal length r of a cyclic binary sequence that contains as substrings all 11-bit words in set R = {binary(x,11):1024≤x≤1990}. (This also ensures numbers<1024? Not automatically for number1023? But if R includes all words 10000000000..11111000110, do their substrings cover every binary representation of m<1024? For m<1024 length l≤10, can we find a required 11-bit word containing m? Need check worst m=1023 ten ones. Required R words have no six consecutive initial ones but may contain ten consecutive ones after an initial zero? An 11-bit required word has length11; to contain ten consecutive ones, it would be either 01111111111 (starts0 value1023, not in R because as 11-bit it starts0 and value1023 <1024; but the word with leading0 is not the representation of a length-11 number) or 11111111110/11111111111 (≥2046) not in R. It cannot contain ten consecutive ones and be≤1990? What about 10111111111 (value1535) contains nine? bits after 10: there are nine ones? 1 0 111111111 = nine ones (positions8..0) total run9, not ten. Length11 with a zero in second bit leaves only9 following bits, so max run after an early zero is9 if zero at bit9? To get ten consecutive ones in an 11-bit word, the zero must be at one end: 01111111111 (starts0) or11111111110 (value2046) or all ones. Required R words start1 and≤1990, so none contain ten consecutive ones. Thus R does NOT cover number1023. However the expansion also must contain 1023 separately. Does containing all length-10 words starting with1 (numbers512..1023) follow from R? For a length-10 word v starting1, extend to length11 by appending a bit. If v≤995, append0 gives ≤1990 required; if v≥996, both extensions ≥1992 missing. For v=996..1023, R does not contain v as prefix. Could v appear inside some required R word not as prefix? For v with ten bits starting1 and value≥996 (i.e. v≥111110001100? let's binary), maybe as suffix of a required word w=b v? If w length11 and suffix=v (≥996), then w = b1024+v. If b=0, w starts0 (not in R); if b=1, w≥2020 >1990. So no. Could v appear crossing inside w with offset not 0 or1? w length11 cannot contain a length10 substring starting at offset1 only (suffix) or offset0 (prefix). So high length-10 words996..1023 are not covered by R. They must appear independently. Note binary representations of numbers512..1023 are required themselves anyway (they are in 1..1990). Our reduction earlier to requiring all length-10 words and some length-11 words stands. The counting lower bound using E=967 length-11 words is valid; the good-start/run argument can be applied to required length-11 words only. It gives z≥194. Additionally, required length-10 high words996..1023 (28 words) are not contained in required length-11 words; but they are contained as length-10 substrings starting at one positions. Do they impose more zeros? They are all words with first five bits11111 and value≥996 (111110001100? let's compute 996 decimal binary length10 =1111100100? Wait 996=512+256+128+64+32+4 = bits9..0:1 1 1 1 1 0 0 1 0 0? Let's recalc: length10 values512..1023. 996-512=484; 484=256+128+64+32+4. Bits9..0: bit9=1 (512), bit8=1(256), bit7=1(128), bit6=1(64), bit5=1(32), bit4=0(16), bit3=0(8), bit2=1(4), bit1=0, bit0=0 =>1111100100. Earlier I said source995 for label1990; source=floor(1990/2)=995. 995 binary=1111100011 (512+256+128+64+32+2+1=995). 996=1111100100. Good. High length-10 words v=996..1023 all start with11111001? Not all; v=1023=1111111111. They have first five bits11111 and the sixth bit? For v≥996, first six bits? v=996 bits first6=111110 (since bit4=0), v=1000..? Some have first6=111110 or111111. The word1023 has first six ones. As length-10 required words, if a length-10 substring starts at position i and is required (number512..1023), does it need a zero within first? For high v with first six? Number1023 ten ones must occur; this forces a run of at least10 ones. Already counted in zeros? The run argument for length-11 required words: each one-run contributes at most5 good starts for length-11 required windows. High length-10 required words like1111111111 can only start in the last? If a run of ones length ℓ, occurrences of ten consecutive ones start at positions j≤ℓ-9 within the run. They are not good for length-11 (unless followed by zero within five bits, impossible if ten ones ahead). They consume one-start positions that cannot be used for required length-11 windows. But for lower bound z≥194, we only used length-11 required windows. If there are runs longer than5, the first ℓ-5 starts in the run are bad for length-11; they may be needed for length-10 high words (e.g. ten ones). This could increase required zeros? Let's see. To contain all length-10 words512..1023, including v with prefix11111 high, some start positions must have runs of ones length at least? Word1023 needs run length≥10. That run contributes only5 good length-11 starts (last five) but has ℓ≥10 one positions; the first ℓ-5≥5 one positions cannot start required length-11 windows. They are "wasted" ones for E. Our inequality E≤5z remains valid (good starts≤5z) regardless of wasted ones. w≥E + wasted? Not necessarily: the same one position cannot start two distinct required substrings of different lengths? It can start a length-10 required word and a length-11 required word simultaneously if the length-11 word is required and its first10 bits also a required number. For high length-10 word1023, the length-11 extension cannot be required, so the start position for1023 is not counted in E. Therefore total ones w must be at least E + s, where s is number of start positions needed exclusively for length-10 required words whose length-11 extensions are not required and which cannot coincide with starts of required length-11 windows. This could increase w and hence r. But maybe a start position for a high length-10 word v=996..1023 can also start a required length-11 word? If the 11-bit window starting there is required (≤1990) and its first10 bits =v. For v=996..1023, appending any bit gives ≥1992, not required. Could the 11-bit window be required while first10 bits v≥996? no. So those start positions are disjoint from the E good starts. How many distinct high length-10 words v=996..1023 (28 words) must appear, each requiring a start position not good for E. But some may be substrings of other required length-10/11 words? They are required themselves; to appear, each needs a start position. Could one start position serve two high length-10 words? A length-10 substring starting at a position is unique, so no. Could a high length-10 word appear starting at a position with bit0? It starts with1, so no. Thus we need at least28 additional one-start positions that are bad for length-11. Are these included in w≥967? The E required length-11 windows need 967 good starts; high length-10 words need 28 bad starts. They are disjoint, so w≥967+28=995. Is that valid? Wait length-10 required words include all512..1023. For v=512..995, we can choose a start position that also starts a required length-11 extension (append appropriate bit ≤1990). For v=512..994, both extensions required; for v=995, extension0 (1990) required. Thus their occurrences can coincide with the E length-11 required windows: each required length-11 window's first10 bits is a length-10 number between512 and995? Let's check required length-11 labels1024..1990 have prefix floor(L/2) =512..995. Yes. So the E=967 length-11 required windows, when truncated to first10 bits, give exactly numbers512..995 (each? L=1024..1990 -> prefixes512..995; since L includes both parities for512..994 and only even for995, the set of prefixes is512..995). Great: numbers512..995 are covered as prefixes of required length-11 windows. Numbers<512 are covered by extending to length10? For x<512, extend to a length-10 word v between512 and995 maybe with appended bits so that v≤995 and then to length11 required. Need for any x<512, choose v length10 starting with binary x and ≤995. If we append zeros to x to length10, v=x2^{10-l}. For x up to? The maximum x<512 is511 (l=9): v=5112=1022 >995. Appending zeros may overshoot. But we can choose a required length-11 word w≤1990 that contains x. Simpler: Our E windows with prefixes512..995 contain as prefixes all length-10 words v=512..995. Do these cover all x<512? For x with length l≤9, append zeros to length10 gives v=x2^{10-l}. For l≤9, minimum for x=1 is512, maximum for x=511 (l=9) is1022. If v≤995, good; if v>995 (x large length9), maybe choose append not zeros? To contain x as prefix of a length-10 v≤995, we need v = x2^{10-l}+y with 0≤y<2^{10-l} and v≤995. Since x2^{10-l} for l≤9: if l≤8, x≤255, v≤2554=1020 but can choose y=0 maybe >995 for x≥249? Hmm not guaranteed. Alternative extend x to length11 directly: w=x2^{11-l}+y. Need 1024≤w≤1990. Since x≥1, choose y=0 gives w0=x2^{11-l}. For l≤10, w0 between1024 (x=1,l? if x length l, x2^{11-l}; for l=1,x=1 ->1024) and for x=511,l=9 ->511*4=2044 >1990. For large x, w0 may exceed1990. Can choose y negative? no. If w0>1990, any extension with same prefix x to length11 is ≥w0>1990. So x cannot be prefix of a required length-11 word. It might appear as a non-prefix substring inside a required length-11 word. For x=511 (111111111, nine ones), is there a required length-11 word ≤1990 containing nine consecutive ones? Yes e.g. 10111111111? That's length11 value? bits 1 0 111111111 (nine ones) =1024+0+256+128+64+32+16+8+4+2+1? Wait after 10 there are nine bits all ones (256+...+1=511), total1535, required. It contains x=111111111 (nine ones) starting at offset2. Good. For x=511, okay. For x with binary length9 and high value, likely can embed after an early zero. We need not detail; the original problem includes all numbers1..1990, and counting length11 required is valid lower bound. For construction, covering shorter numbers may be nontrivial but often follows if all length-11 words up to1990 plus high length-10 words are present? Let's not get lost; lower bound can be strengthened by considering numbers996..1023 maybe.

Are numbers996..1023 already included in the list1..1990; yes. Their binary representations length10. They are not prefixes of any required length-11 word (since doubling ≥1992). Could they be suffixes or internal substrings of required length-11 words? As length10 substrings of an 11-bit word, offset0 prefix or offset1 suffix. For v≥996, suffix case w=b v: if b=0, w as 11-bit starts0 and numeric value v (<1024), not one of the length-11 representations (which have no leading zero and value≥1024). But the expansion could contain w with leading0 as a length-11 window; the required length-10 v would appear as its suffix. However that length-11 window starts with0, so it is an extra non-required window. It still requires a start position (with bit0) for the window, but the substring v starts at the next bit (a one). The start position of v itself is a one; can it also start a required length-11 window? It starts with v≥996, so no. Thus v's starting one is disjoint from good starts. So indeed each v=996..1023 needs a distinct one-start that is bad for length-11. There are28 such numbers. Wait v=995 binary1111100011: appending0 gives1990 required, so its start can be good. v=996 appending0=1992 missing; but maybe v=996 appears as suffix of a required length-11 word w=0 1111100100 (starts0, value996) plus? Not required length11. Or as substring of a longer required word length>11? Required representations max length11 (1990). There are no required words longer than11. But the expansion infinite can contain v across boundaries; the start position of v is what matters. If the bit before v is0 and bits after arbitrary, the length-11 window starting at v (v plus next bit) is missing, and the length-11 window ending at v (previous bit plus v) starts with0 if previous bit0 or missing if previous bit1. Not relevant. So w≥967+28=995. Then with z≥194, r≥995+194=1189. But can the 28 high length-10 words share start positions with each other via different lengths? No same length distinct words need distinct starts. Good. What about numbers1008..1023 maybe have first six bits111111? They require runs etc. But count28 stands.

However, are numbers996..1023 not already contained in the binary representations of larger required numbers 1984..1990? Those length11 words have prefix11111 and suffix000000..000110. Their length-10 substrings: prefix for w=1984+s (s=0..6) is floor(w/2)=992+floor(s/2)? Let's compute w=11111000000 (1984) prefix length10 =1111100000 (992), suffix=1111000000 (960). w up to1990 prefix=995, suffix? The length-10 substrings of w=11111000 s(3 bits?) are around960..995, not996..1023. So no.

Let's verify threshold: Required length-11 words L=1024..1990. Their prefixes floor(L/2) range512..995. Thus length-10 numbers512..995 are covered as prefixes. Numbers996..1023 (28 numbers) are not prefixes; nor suffixes because suffix = L mod1024 ranges0..966. Actually L mod1024 ranges0..966; high v≥996 not suffix either. Internal length10 in length11 only prefix/suffix. So they are disjoint. Good. Then w≥967+28=995. But do numbers996..1023 require a zero within first? They are length10; for v=1023 all ones no zero. They occur starting at one positions. These starts are bad for length-11 (since any length-11 window starting there begins with v≥996 plus bit -> ≥1992, missing). They also may be in one-runs. The run contribution argument: In a one-run of lengthℓ, at most5 starts are good for length-11 required windows. The high length-10 words require starts with certain patterns; in particular word1023 requires a run length≥10 and a start not among last? It could be one of the bad starts. Count w≥995 is valid.

Then z≥ceil(967/5)=194 (from length-11 good starts). So r=w+z≥995+194=1189. Then n≥r+1≥1190; n odd≥1191. Candidate? 1191=3397? 1+1+9+1=12 yes composite. Need find smallest odd n with ord_n(2)≥1189 and possible. Since for prime p, need p-1≥1189 -> p≥1190; smallest odd prime ≥1191? 1191 composite (3397), 1193? Check prime sqrt34.5. 1193 mod3=14?1+1+9+3=14 no; mod5 no;7:7170=1190 rem3;11108=1188 rem5;1391=1183 rem10;1770=1190 rem3;1962=1178 rem15;2351=1173 rem20;2941=1189 rem4;3138=1178 rem15. prime? yes. p-1=1192=8149? 1192/8=149. Divisors:1,2,4,8,149,298,596,1192. Only divisor≥1189 is1192. Need ord_1193(2)=1192. p mod8 =1 (1192 divisible by8? 8149=1192, p≡1), so 2 is quadratic residue, order divides596 <1189. Not candidate. Next odd numbers: 1195=5239; φ=4238=952 <1189 no. 1197=3399? φ maybe <. 1199=11109? φ=10108=1080 <1189. 1201? prime? sqrt34.6. Test 1201 mod3=4 no;5 no;7 7171=1197 rem4;11109=1199 rem2;1392=1196 rem5;1770=1190 rem11;1963=1197 rem4;2352=1196 rem5;2941=1189 rem12;3138=1178 rem23. prime. p-1=1200=2^435^2. Need order≥1189; divisors of1200 that are≥1189: only1200? Proper divisors≤600 (since largest proper divisor=600). Yes need primitive. p=1201 mod8 =1 (1200 divisible by8), so 2 residue -> order divides600 <1189. Not candidate. 1203 composite (3401) φ=800; 1205=5241 φ=960; 1207=1771? 1771=1207 φ=1670=1120 <1189; 1209=31331? φ smaller; 1211? prime? sqrt34.8. Test mod3=5 no;5 no;7 7173=1211? 7173=1211 yes composite. 1213? prime? mod3=7 no;5 no;7 7173=1211 rem2;11110=1210 rem3;1393=1209 rem4;1771=1207 rem6;1963=1197 rem16;2352=1196 rem17;2941=1189 rem24;3139=1209 rem4. prime. p-1=1212=43101? =2^23101. Largest proper divisor=606 <1189, so need primitive order1212. p mod8: 1212 mod8=4, p≡5 mod8, so 2 nonresidue -> 2^{606}≡-1 (good for factor2? p-1 has 2^2; if 2^{606}=-1 then order has factor4? Since 606=(p-1)/2, yes not divide606; but p-1 only 4303, to have full factor4 need 2^{303} not +1? Actually if 2^{606}=-1, then 2^{303} is a square root of -1; order does not divide606 and has 2-adic valuation2 (full) automatically? If order d divides1212 and not divide606, then v2(d)=v2(1212)=2 yes good). Need check prime factors q=3 and101: 2^{1212/3}=2^{404}≠1 and 2^{1212/101}=2^{12}=4096 mod1213 =4096-31213=4096-3639=457 ≠1. Need compute2^{404}. Could be. If primitive, r=1212. Then n=1213 candidate. For n=1213 (>1024,<2048), repetend bit length = r-10? q≈2^{1212}/1213; ratio to2^{1202}:1024/1213=0.84419; bit length1202, leading zeros10. r-E_extra? If required w≥995, z≥194, r=1212 gives slack r-(995+194)=23. More feasible than1163. But still need prove existence for n=1213. This is not neat; answer maybe not this.

I suspect I'm overcomplicating. Let's look for the intended elegant solution.

Maybe the problem is from a list where answer uses the repeating period of 1/n must be at least 2048 because the binary representations of 1..1990 include 1990 numbers whose total length? No, containing as substrings can overlap. But maybe "包含 1,2,...,1990 各个数字的二进制表示" means the decimal (fraction) representation contains them in that order? The Chinese phrase "能够包含 1, 2, 3, ... , 1990 各个数字的二进制表示" Could be read: contains the binary representations of the numbers 1,2,...,1990 (not necessarily in order). If it meant in order, it would say "依次包含". Not.

If they must appear in order 1,2,3,...,1990 as consecutive substrings (overlaps allowed? probably not), then the required concatenated string length = sum bit lengths = for lengths: length1:1 (1), length2:2 numbers? Let's compute total bits of concatenation of binary reps 1..1990. Number with length L: for L=1:1 (1), L=2:2 (2,3) count2, ..., L=10: count512, L=11: numbers1024..1990 count967. Total bits =11 +22+34+...+10512+11967. Sum_{L=1}^{10} L 2^{L-1} + 11967. Formula sum_{L=1}^{m} L2^{L-1} = (m-1)2^m+1. For m=10:9*1024+1=9217. Plus 10637 =19854. If the expansion of 1/n has period r, to contain a specific string of length19854 (concatenation) need r≥19854 maybe; n≥19855. Not likely? 1990 chosen maybe because total length ~? The concatenation length 19854 close to? no.

Maybe "包含 1,2,...,1990 各个数字的二进制表示" means the fractional part contains all the digits that appear in the binary representations? That is just digits0 and1. Then n=2 works (0.1) contains1 and0? Too trivial. No.

Let's search memory snippets: I remember an IMO Shortlist 1990 problem (maybe C5?) about binary expansion of 1/n containing all numbers up to 1990. The solution answer might be "n=2^{11}-1=2047"? There is a known concept: If n=2^k-1, then 1/n has period k with repetend 000...001. Multiples? Not. If n=2^k+1 (Fermat), then 1/n has period 2k with repetend 000...01000...0? For n=2^{11}+1=2049 (not prime), period? Since 2^{11}≡-1 mod n if n=2049? But n=2^{11}+1 exactly, gcd? 2^{11}≡-1 mod2049, so ord=22. The expansion 1/2049 =? Using 1/(2^{11}+1)=2^{-11}/(1+2^{-11})=2^{-11}-2^{-22}+2^{-33}-... This has binary digits with alternating carries, not simple. But n=2049 is the smallest n with 2^{11}≡-1? It may have period22. Not enough for substrings.

Wait maybe the problem asks for smallest n such that the period of 1/n (as a repeating block) contains the binary representations of 1..1990 as factors when the block is read in all rotations. For n=2^m-1, period block is 0^{m-1}1. Not. For n=2^m+1, repetend maybe 0^{m-1}10^{m-1}? Let's compute 1/(2^m+1). Since (2^m+1)(2^m-1)=2^{2m}-1, 1/(2^m+1)=(2^m-1)/(2^{2m}-1), repetend length2m is binary of 2^m-1 = 0^m 1^m? Specifically (2^{2m}-1)/(2^m+1)=2^m-1, which in 2m bits is 0...0 (m zeros) followed by m ones. Ah! For n=2^m+1, repetend = 0^m 1^m. That contains only zeros then ones; not all.

For n=2^m-1, repetend =0^{m-1}1. For n=(2^{2m}-1)/(some de Bruijn integer), arbitrary.

Maybe the smallest n is obtained by taking m=11 and n=2^{11}-1990? Let's see. There is a connection: We want 1/n expansion to include binary rep of k for k≤1990. For odd n and period r, the repetend integer q=(2^r-1)/n. The length-r repetend contains a given word w iff there exists j such that w appears in q rotated. This is equivalent to existence of integer a such that fractional part {a/n}? Hmm. Multiples k/n have repetend kq mod(2^r-1). But we only have 1/n.

Alternative number theory: The binary expansion of 1/n contains the binary representation of m iff there exists integers a,b such that m/2^t ≤ {2^a/n} < (m+1)/2^t (where t=bitlength(m)), i.e. there exists a with fractional part of 2^a/n in interval I_m=[m/2^t,(m+1)/2^t). Since {2^a/n} = (2^a mod n)/n. So the condition is: the set of residues {2^a mod n} (a≥0) intersects every interval (n m/2^t, n(m+1)/2^t) for m=1..1990. Because 2^a mod n runs over the multiplicative semigroup generated by2 modulo n (eventually cycles through subgroup <2> in units times maybe). If n is odd, residues are powers of2 mod n in the unit group. We need for each m, some power residue x=2^a mod n falls in interval J_m=(ceil? ) corresponding to prefix m: x/n ∈ [m/2^t,(m+1)/2^t). That is n m/2^t ≤ x < n(m+1)/2^t. The length of this interval is n/2^t. For m around1990, t=11, length n/2048. To guarantee hit by powers maybe if interval length≥1? Not guarantee. If n≥2048, each interval length≥1, so there is an integer x in every such interval, but x must also be a power of2 mod n. If n=2^r-1? Powers of2 mod n are just 1,2,4,...,2^{r-1}; sparse. Not.

If n>2048, the intervals J_m for m≤1990,t≤11 have length n/2^t ≥ n/2048 >1, so each contains at least one integer; but we need a power of2 modulo n. If 2 is a primitive root modulo n and n-1 is large, the powers are all residues 1..n-1 (if n prime). Then certainly each interval containing an integer residue is hit. Aha! If n is prime and 2 is a primitive root modulo n, then the residues 2^a mod n run through all nonzero residues mod n. Therefore for any interval of real numbers of length >1 within (0,n), it contains an integer residue r, and there is a power 2^a≡r mod n; hence the corresponding binary word appears. This gives a sufficient condition: choose prime n such that 2 is primitive mod n and n/2048 ≥1? More precisely for every m≤1990 with bitlength t≤11, the interval [n m/2^t, n(m+1)/2^t) has length n/2^t ≥ n/2048. If n≥2048, length≥1. Any half-open interval of length≥1 contains an integer; if length=1 exactly may fail depending endpoints, but choose n>2048 to have length>1. Then it contains an integer between1 and n-1? Need ensure. For m≥1,t≤11, lower endpoint n m/2^t >0. Upper endpoint n(m+1)/2^t ≤ n1991/2048 < n (since1991<2048). For m=1990, upper=n1991/2048<n. Good. If length>1, the interval contains an integer x with 1≤x≤n-1. If n is prime and 2 primitive mod n, x is a power of2 modulo n, so word m appears. Thus any prime n>2048 with 2 primitive mod n works. The smallest such prime? Candidate 2053? We checked 2051 composite, 2053 prime maybe and p≡5 mod8 (2 nonresidue). Need test primitivity. p-1=2052=2^23^319 (since2052/4=513=27*19). To be primitive, need 2^{1026}≠1 (nonresidue gives -1), 2^{684}≠1 (factor3), 2^{108}≠1 (factor19? since2052/19=108), and also for factor3^3? The prime divisors of p-1 are2,3,19. Need check 2^{2052/3}=2^{684}≠1 and 2^{2052/19}=2^{108}≠1. Nonresidue handles factor2. We can perhaps compute using relation? 2053 = 2048+5 =2^{11}+5. Nice! This is likely intended! n=2053 = 2^{11}+5. Since 1990<2048, the sufficient prime just above2048 with a simple form 2^{11}+5. Is 2053 prime and 2 primitive? Let's test with relation x=2^{11}=2048≡-5 mod2053. Great! We can compute powers elegantly. If 2^{11}≡-5, then we can check primitivity.

p=2053, p-1=2052=11? Wait 2052 = 2^{11}+4? Actually 2053-1=2052=4513=42719 =2^23^3*19. It does NOT have factor11. But relation 2^{11}≡-5.

Check 2 is primitive mod2053:

Now minimal? We need prove no smaller n works. The sufficient construction gives n=2053. Need lower bound n≥2053? How to show elegantly? Perhaps using the binary representation of 1990 and all numbers? We need show n must be >2048. Is that true? Let's test with our earlier lower bounds: n≥1191 maybe, far below2053. So cannot prove n≥2053 from substring counts unless there is an additional theorem: For 1/n to contain all binary words up to length11, n must exceed2048? Is that plausible? Let's test with a hypothetical prime p=1213 with 2 primitive. Then residues powers of2 mod p run through all nonzero residues. For a word m of length t, the interval length p/2^t. For t=11, length p/2048≈0.592 <1. Some intervals may contain no integer residue, so primitivity does not guarantee. But maybe because the intervals for m=1..1990 (with variable t) are spaced such that every interval of length p/2048 <1 could still be hit by powers (all residues) if the interval happens to contain an integer. Since powers are all residues mod p (if primitive), the condition for word m is simply that the interval J_m contains an integer residue. If p<2048, for t=11 intervals length<1; some contain integers, some not. To contain all m=1024..1990, we need for each m, the fractional interval [p m/2048, p(m+1)/2048) contains an integer. Equivalently there exists integer x with x/ p? Let's derive: x = power residue between p m/2048 and p(m+1)/2048. This means floor? There must be integer x in that interval. For m and m+1 consecutive, intervals tile [p1024/2048=p/2, p1991/2048) into 967 consecutive intervals each of length p/2048<1. For each of 967 consecutive intervals to contain an integer, since length<1, each interval must contain a distinct integer and there can be no gaps. The integers in consecutive intervals would have to be spaced <1 apart, impossible for many consecutive intervals unless length close to1? Let's examine. For m=1024..1990, intervals I_m=[p m/2048, p(m+1)/2048). They are consecutive (end of one = start of next) covering [p/2, p1991/2048) of total length p(967/2048) ≈0.472p. If each interval contains an integer, because intervals are disjoint and ordered, the chosen integers x_m must be strictly increasing with x_{m+1}-x_m ≥1. But the start points increase by p/2048. If p/2048<1, can each interval still contain an integer? Yes if the fractional parts align: e.g. length0.6 intervals can each contain an integer only for limited runs; over 967 consecutive intervals, the integers would need to increase by roughly0.6 per step, but integer differences are at least1, causing the chosen integer to drift ahead; eventually an interval may contain no integer. More precisely, if each I_m contains an integer x_m, then x_{m+1}≥x_m+1. But the upper bound for x_m is < p(m+1)/2048. Thus x_m < p(m+1)/2048 and x_m ≥ x_{1024}+(m-1024) ≥ ceil(p/2)+(m-1024). For m=1990, need an integer x_{1990} ≥ ceil(p/2)+966 and < p1991/2048. This implies p1991/2048 > p/2 +966 (approx) -> p*(1991/2048 -1/2)=p*(1991-1024)/2048 = p967/2048 >966. So p > 9662048/967 ≈2045.89. Aha! This gives p≥2046 maybe. Nice! Let's formalize. If each interval I_m (m=1024,...,1990) contains an integer x_m, then x_{m+1}>x_m (distinct ordered) hence x_{m+1}≥x_m+1. There are967 integers. The first x_{1024} ≥ floor? I_{1024}=[p/2, p/2+p/2048). The smallest integer in it is at least ⌈p/2⌉ (if p even, p/2 included; but p odd maybe ceil). The last x_{1990} ≥ ⌈p/2⌉+966. Also x_{1990} < p1991/2048. Therefore ⌈p/2⌉+966 < p1991/2048. For odd p, ceil(p/2)=(p+1)/2. Then (p+1)/2+966 < p1991/2048. Multiply 2048: 1024(p+1)+9662048 <1991 p. Left =1024p+1024+1,978,368? Compute9662048 =2048(1000-34)=2,048,000-69,632=1,978,368. +1024 =1,979,392. Inequality:1024p+1,979,392 <1991p ->1,979,392 <967p -> p >1,979,392/967. Compute 9672047 =? 2048967 -967 =1,979,136 -967=1,978,169. Remainder1,223. 9672048=1,979,136. Our numerator1,979,392 =9672048 +256. So p >2048 +256/967 =2048.2647. Thus p≥2049 (if integer). Nice! Since n odd, n≥2049. This is close. Then we can rule out n=2049,2051 and show2053 works. This seems plausible and elegant! Great.

Let's refine lower bound for general n (not necessarily prime). We don't need powers all residues. For a fixed m with t=11 (m=1024..1990), if binary representation of m appears in 1/n, then there exists a nonnegative integer a such that fractional part {2^a/n} lies in I_m=[m/2048,(m+1)/2048). Multiply by n: there exists integer y= n{2^a/n}. If n is odd, y = 2^a mod n (in {1,...,n-1}) because 2^a not divisible by n. If n even, write n=2^s u odd; for a≥s, fractional part corresponds to residue modulo u times power? But minimal n likely odd; we need handle even n in lower bound. We can first show n must be odd: If n=2^s u with u odd. The binary expansion of 1/n is just that of 1/u shifted s places to the right (with s initial zeros). Any occurrence of the binary representation of an integer m≥1024 (which begins with1) cannot use those initial zeros, so it must already occur in 1/u. Hence if n works, its odd part u works and u≤n. Therefore a minimal n must be odd. Good. For odd n, residues y_a=2^a mod n are integers in1..n-1 (since n odd >1; if residue0 impossible). If word m appears, there is a with y_a/n ∈ I_m, i.e. y_a ∈ J_m=[ n m/2048, n(m+1)/2048 ). J_m must contain an integer (the residue). Good. We don't need primitivity for lower bound; just each J_m contains at least one integer (the residue). Then the counting/drift argument applies to n. Let's set n odd. For m=1024,...,1990, let x_m be an integer in J_m. Since intervals J_m are disjoint and increasing, x_{1024}<x_{1025}<...<x_{1990}; integers, so x_{m+1}≥x_m+1. Thus x_{1990}≥x_{1024}+966. Now x_{1024}≥⌈ n1024/2048⌉=⌈n/2⌉=(n+1)/2 (n odd). And x_{1990}< n1991/2048. Hence (n+1)/2 +966 < 1991 n/2048. Solve as above: Multiply by2048:1024(n+1)+966·2048 <1991n. 966·2048=1,978,368; plus1024 =1,979,392. So 1024n+1,979,392 <1991n ⇒1,979,392 <967n ⇒ n >1,979,392/967 =2048 +256/967. Therefore n≥2049. Since n odd, n∈{2049,2051,...}. Good.

Need rule out n=2049 and n=2051 (and maybe n=2050 even but minimal odd argument handles: if even n worked, odd part≤n; lower bound applies to odd part, so odd part≥2049; even n> odd part could be2050 with odd part1025? impossible by lower bound; but simply since minimal n odd, candidates ≥2049 odd). Check n=2049=3·683. Need show it does NOT work. We can use order/period: ord_{2049}(2)=lcm(ord_3(2)=2, ord_683(2)). Since 683 divides2^{11}+1? Actually 2049=3*683 and 2^{11}=2048≡-1 mod683, so ord_683(2)=22 (need ensure not divides11; 2^{11}≡-1 not1, and if ord d|22 and d∤11 then d=22; proper divisors of22 are1,2,11; 2^1=2,2^2=4,2^{11}=-1, so d=22). lcm(2,22)=22. Thus the binary expansion of1/2049 has period22 (after no preperiod; n odd). It has only22 distinct substrings of length11, but needs967 distinct (numbers1024..1990). Impossible. Good.

n=2051= ? Factor. We found 2051=7293 (7293=2051). ord_7(2)=3 (since2^3=8≡1). ord_293(2)? 292=473. Need upper bound λ(2051)=lcm(6,292)=292? Actually φ(7)=6, φ(293)=292, lcm=292. Thus period r≤292 <967. Impossible (needs967 distinct length-11 substrings). Simpler: ord_{2051}(2)≤lcm(6,292)=292<967. Good. Need ensure factorization 2051=7293 (293 prime). 293 prime (sqrt17.1; not divisible by3,5,7 (741=287 rem6),1126=286 rem7,1322=286 rem7,1717=289 rem4). Good. So n=2051 cannot.

Then n=2053 works if prime and 2 primitive. Need prove 2053 prime and 2 primitive. Prime check: sqrt2053≈45.3; test primes up to43. We need present concise. Divisibility by primes p≤43: 3 (sum10 no),5 no,7:7293=2051 remainder2,11:11186=2046 rem7 (or 11187=2057 rem-4),13:13158=2054 rem-1,17:17120=2040 rem13,19:19108=2052 rem1,23:2389=2047 rem6,29:2970=2030 rem23,31:3166=2046 rem7,37:3755=2035 rem18,41:4150=2050 rem3,43:4347=2021 rem32. So prime.

Primitivity of2 mod2053: p-1=2052=2^2·3^3·19. It suffices to show for q=2,3,19, 2^{2052/q} not ≡1. We can use computations. Need present correct modular computations. Let's verify our computations carefully, maybe use exact manual or perhaps use a small amount of mental arithmetic. Since user forbids programming, but as model we can compute. Let's verify thoroughly.

Given p=2053 and 2^{11}=2048≡ -5 (mod2053).

Alternatively compute 2^{684} via repeated squaring of2 maybe smaller? Relation with5 is fine.

Thus 2^{1026}≡-1, 2^{684}≡1855, 2^{108}≡1929, none1; by the standard criterion, ord_{2053}(2)=2052. In particular 2 is primitive root modulo prime2053, so the residues 2^a mod2053 (a=0,...,2051) run through all integers1,...,2052.

Then sufficient proof: For any integer k with1≤k≤1990, let t be its binary length (2^{t-1}≤k<2^t; t≤11). Consider interval I_k=[2053 k/2^t, 2053(k+1)/2^t). Its length is2053/2^t ≥2053/2048 >1. Also it lies inside(0,2053): lower>0; upper≤2053·1991/2048<2053 (since for k≤1990 and t=11 maximal upper; for smaller t, k+1≤2^t so upper≤2053; actually if k=2^t-1 (max for length t) and t<11, upper=2053, endpoint excluded. Example k=1023,t=10: upper=20531024/1024=2053, interval [20531023/1024,2053). It lies within (0,2053), upper excluded =2053. Fine; lower>0. Length>1 so contains an integer r with1≤r≤2052. Need check for k=1023,t=10 length2053/1024>2, contains integer<2053 yes. For k=1,t=1 interval [2053/2,2053) length1026.5 contains integers. Good.) Since length>1, there is an integer r in I_k. Because 0<r<2053 and 2 is primitive mod2053, r≡2^a mod2053 for some a. Then the fractional part {2^a/2053}=r/2053 lies in [k/2^t,(k+1)/2^t), which exactly means the t binary digits after the binary point starting at position a+1 are the binary representation of k. Need handle if 2^a/2053 >1; fractional part. Since a≥0, {2^a/2053}= (2^a mod2053)/2053 =r/2053. Multiplying by2^t: k≤ r 2^t/2053 < k+1, so floor(2^t {2^a/2053})=k. Thus the next t bits are bin(k). Good.

Now lower bound details: We need first prove minimal n odd. Let n=2^s u, u odd. If u=1, expansion terminates and cannot contain e.g.3; so u>1. The binary expansion of1/n is obtained from that of1/u by shifting s places to the right, i.e. adding s zeros after the point. Since every required binary representation begins with1, any occurrence in1/n is entirely after those s zeros; hence1/u also works. As u≤n, a least n must have s=0; so n odd. Good.

For odd n>1, suppose the binary representation of an integer m with 10≤? Actually for m=1024..1990 (length11) appears. Then there exists a≥0 such that m/2048 ≤ {2^a/n} < (m+1)/2048. Let x_m=n{2^a/n}. Because n odd, x_m is the least positive residue of2^a mod n, hence an integer in [1,n-1]. It lies in [nm/2048, n(m+1)/2048). (If {2^a/n}=0 impossible since n∤2^a.) So for each m=1024,...,1990 the interval J_m contains an integer x_m. These intervals are disjoint and consecutive, so x_{1024}<...<x_{1990}; hence x_{1990}≥x_{1024}+966. But x_{1024}≥⌈n/2⌉=(n+1)/2 (n odd) and x_{1990}<1991n/2048. Combine to get n>2048+256/967, so n≥2049; as n odd, n≥2049 and the next possibilities below2053 are2049,2051.

Wait the inequality gives n≥2049 (integer). Since n odd, n≥2049. We then exclude2049,2051. Good. Check algebra with strict/non-strict endpoints: x_{1024}≥ceil(n/2). If n odd, n/2 = k.5, ceil=(n+1)/2, and interval J_{1024}=[n/2, n/2+n/2048). Includes lower endpoint. x integer ≥(n+1)/2. Good. x_{1990}<1991n/2048 strict. x_{1990}≥x_{1024}+966 (there are966 increments from1024 to1990: 1990-1024=966). So (n+1)/2+966 ≤ x_{1990} <1991n/2048. Actually we have ≤ then <: (n+1)/2+966 ≤ x_{1990} < ...; combine to (n+1)/2+966 < ... (since if left≤x and x<right, left<right). Good. Solve carefully: (n+1)/2+966 < 1991n/2048. Left = n/2+1/2+966 = n/2+966.5. Multiply2048:1024n + (966.52048). 966.52048 =9662048+1024 =1,978,368+1,024=1,979,392. Right1991n. So 1024n+1,979,392 <1991n ->1,979,392<967n. Divide: 9672048 =? Compute 2048*(1000-33)=2,048,000 -67,584 =1,980,416? Wait earlier I may have miscomputed 9672048! Let's recalc carefully! 2048967 =2048*(1000-33) =2,048,000 -67,584 =1,980,416. Oh no earlier I wrote1,979,136 (off by1280). Let's recalc 9662048 and numerator. 2048966 =2048*(1000-34)=2,048,000 -69,632 =1,978,368. Correct. Add1024 =1,979,392. Now 9672048 =2048(966+1)=1,978,368+2,048=1,980,416. Yes. Numerator1,979,392 is less than9672048 by1,024. Wait then inequality1,979,392 <967n. For n=2048, RHS=1,980,416, so inequality holds (1,979,392<1,980,416)! That would give n> ? Compute 1,979,392/967. Since9672047 =9672048 -967 =1,980,416-967=1,979,449. Numerator1,979,392 is 57 less than9672047. Ah! Then 1,979,392/967 =2047 -57/967 ≈2046.941. So inequality gives n>2046.94, i.e. n≥2047; with n odd n≥2047. Not enough. I made an arithmetic error earlier (967*2048). Let's recalc thoroughly. This changes lower bound! We need check. Left constant maybe wrong? Let's derive from scratch.

Intervals for m=1024..1990. Count =967. If each contains integer x_m increasing, then x_{1990} ≥ x_{1024}+(967-1)=x_{1024}+966. Good. x_{1024} ≥ ceil(n1024/2048)=ceil(n/2). For n odd = (n+1)/2. x_{1990} < n1991/2048. Inequality: (n+1)/2 +966 <1991n/2048. Multiply2048: (n+1)1024 +9662048 <1991n. (n+1)1024=1024n+1024. 9662048=1,978,368. Sum constant=1,979,392. Inequality:1024n+1,979,392 <1991n →1,979,392 <967n. Now divide: 9672047 =? 2047(1000-33)=2,047,000 -67,551 =1,979,449. Yes numerator is 1,979,392, which is 57 less than 9672047. So n >2047 -57/967 =2046.941. Thus n≥2047. Since n odd, n≥2047. This lower bound only excludes up to2045. We need exclude 2047,2049,2051 perhaps. Candidate sufficient2053. Need check n=2047,2049,2051. 2047=2^{11}-1=2389? Actually2047=2389. ord? λ=lcm(ord23=11? 2^11=2048≡1 mod23? 2389=2047, 2^11≡1 mod2047 actually since n=2^11-1, ord_n(2)=11). Period11, impossible. 2049 period22, 2051 period≤292. Good. So lower bound n≥2047 and exclude2047/2049/2051. Nice. But wait n=2047 works? It has period11, only 11 distinct length11 substrings; needs967, no. Good.

This is much more coherent. The arithmetic constant: Since numerator <9672047, lower bound n≥2047. If numerator had been slightly bigger, we'd get2049. Let's verify x_{1024} lower maybe can be improved: For n odd, J_{1024}=[n/2, n/2+n/2048). The smallest integer in this interval is not necessarily ceil(n/2)=(n+1)/2; but is it guaranteed to contain an integer? If the expansion works, yes. But for lower bound, x_{1024} is that integer. It could be as low as (n+1)/2. Good. Could x_{1024} actually be greater because interval length n/2048. For n near2047, length≈1, so J_{1024}=[1023.5,1024.4995) for n=2047? It contains integer1024. Our lower (n+1)/2=1024 exactly. Good. x_{1990}<20471991/2048. Compute n=2047: upper=1991 -1991/2048 ≈1990.0278, so x_{1990}≤1990. But x_{1024}+966=1024+966=1990. This is consistent: x_{1990}=1990 possible. So counting doesn't exclude2047. Good. For n=2049: J1024=[1024.5,1025.5) contains1025; x1024≥1025; +966=1991; upper=2049*1991/2048 =1991+1991/2048≈1991.972; x1990<1991.972, integer≤1991. So x1990=1991 possible. Not excluded. Need period argument for2049. For n=2051 similar possible. Good.

But hold on: For n=2047, J1024=[1023.5,1024.4995) contains x=1024; x1990 upper=1990.027, so x1990≤1990; x1024+966=1990 exactly. Thus the counting argument forces for n=2047 that x_m= m? Specifically x_{1024}=1024 and x_{m}=m for all m=1024..1990 (since must increase by exactly1 and end≤1990). That would mean for each m, residue x=m is a power of2 modulo2047. But modulo2047=2^{11}-1, powers of2 are only1,2,4,...,1024 (and repeat). For m=1025, x=1025 would need be a power of2 mod2047; but powers mod2047 are ≤1024 (2^0..2^10) and then repeat; 1025≡? 2^{11}≡1, so residues are {1,2,4,...,1024}; 1025 not in set. This is another way to exclude. Period11 simpler.

For n=2049 (3*683) period22; not enough. For n=2051 period≤292. Good.

But what about n=2047 lower bound: period r=ord_{2047}(2)=11 because 2^{11}≡1 mod2047 and no smaller (2^1..2^10 not1 mod2047? Since2047>2^10+? 2^{10}=1024 not1; yes ord=11). Needs967 length11 substrings; impossible. Good.

Now, is n=2053 indeed the smallest odd after exclusions? Odd numbers ≥2047 and <2053: 2047,2049,2051. Yes. Even numbers? Minimal n odd proof covers. Good.

This seems like the intended solution! The key was intervals for m=1024..1990 and the count gives n≥2047 (not huge), then check three Mersenne-ish candidates, and n=2053=2^{11}+5 works by primitivity. Nice. Let's make sure the sufficient condition for n=2053: It relies on 2 being primitive root modulo2053 so all residues appear as 2^a mod n. For a prime p, if 2 is primitive, the sequence 2^a mod p for a=0,...,p-2 covers all nonzero residues. The binary expansion of1/p has period p-1=2052. Does it then contain every possible word whose interval length>1? Yes because for any k≤1990 length t≤11, interval I_k length p/2^t ≥p/2048>1 and lies within (0,p) (except possibly upper=p for k=2^t-1,t<11; excluded). It contains an integer residue r. Then some power residue equals r. Good. Need note if interval length>1 guarantees an integer r with 0<r<p. If upper=p (excluded) and lower<p-? length>1, still contains integer ≤p-1. For k=2^t-1 with t≤10, interval=[p(1-2^{-t}), p). Length p/2^t ≥2053/1024>2, so contains integer <p. For k=1990,t=11 upper=p*1991/2048 <p. Good.

Let's verify p=2053/2048>1 by5/2048; length for t=11 =1.00244140625 >1. Good. An interval of length just over1 always contains an integer? Any half-open interval [a,a+L) with L>1 contains an integer: Let m=ceil(a). If m<a+L? Since ceil(a)<a+1 (unless a integer? If a integer, ceil=a <a+1; if a noninteger, ceil<a+1). Because L>1, a+1<a+L, so m<a+1<a+L. Yes. Good.

Now prove 2053 prime and primitive. Let's see if there is an even more elegant way using n=2053=2048+5. For primitivity, p-1=2052=4513=43^3*19. Need show order not dividing (p-1)/q for q=2,3,19. We can present the computations compactly. Let's double-check 2^{684} and2^{108} via an independent method to avoid a computational error in final.

Use relation a=2^{11}≡-5. For 2^{684}: exponent684 = 1162 +2. Yes. (-5)^{62}=5^{62}. We computed5^62=977. Let's verify by using repeated squaring of2 directly maybe with smaller residues: 2^{11}=-5. 2^{22}=25. 2^{44}=625. 2^{88}=625^2≡555 (same as5^8? yes 2^{88}=(2^{11})^8≡5^8=555). 2^{176}=555^2≡75. 2^{352}=75^2≡1519. We need2^{684}=2^{352+176+88+44+22+2}? 352+176=528; +88=616; +44=660; +22=682; +2=684. Multiply residues:151975=113925 rem1010; *555 rem81; *625 rem1353; *25 rem? Wait after *625 we got1353 (that's exponent? Let's map: Start 2^352=1519; *2^176=75 ->2^528=1010; *2^88=555 ->2^616=81; *2^44=625 ->2^660=1353; 2^22=25 ->2^682 =135325=33825 rem977; 2^2=4 ->2^684=3908 rem1855. Good. This presentation uses powers of2 directly and is clean: 2^{11}≡-5; hence 2^{22}≡25, 2^{44}≡625, 2^{88}≡555, 2^{176}≡75, 2^{352}≡1519; then combine. Wait 2^{22}=(2^{11})^2≡25; 2^{44}≡625; 2^{88}≡625^2≡555; yes. Good. Then 2^{684}=2^{352}2^{176}2^{88}2^{44}2^{22}2^2 ≡1519·75·555·625·25·4. We can state result1855. To be transparent, show intermediate after first three:1519·75·555 ≡81? Let's compute151975=1010; *555=81 (mod). Then 81·625·25·4. 625·25·4=62500; 81=5,062,500. Mod2053? Our previous after81625=1353; *25=977; *4=1855. Good. We can just give final.

For 2^{108}: 108=88+11+9? Or use powers above: 108=88+11+9 (since2^88≡555,2^11≡-5,2^9=512). Product555*(-5)512 = -2775512. -2775 mod2053 = -722 =1331; 512=1929. Alternatively 108=44+44+11+9? Not. Use 2^{108}=2^{88}2^{11}2^9≡555·(-5)·512≡1929. Check555(-5)=-2775≡-722≡1331; *512≡1929. Good.

For 2^{1026}: Could compute using p≡5 mod8. If audience maybe not know quadratic residue of2; we can also compute via relation:1026=352+352+176+88+44+11+3? Let's see 3522=704; +176=880; +88=968; +44=1012; +11=1023; +3=1026. Residues:2^352=1519; squared? We can compute2^{1026}=2^{1023}8. Since1023=1193, 2^{1023}=(2^{11})^{93}≡(-5)^{93}= -5^{93} (odd). Need5^{93}; we have5^62=977; 5^32=1519; 5^93=5^625^32? 62+32=94, no. 5^{93}=5^{64}5^{16}5^85^45^1? Need more. Euler criterion simpler. Since p≡5 mod8, 2^{(p-1)/2}≡-1. We'll state this known fact. For a self-contained math solution, it's acceptable. If we want avoid Legendre symbol, we can prove 2^{1026}≡-1 by noting from computations maybe 2^{684},2^{108} not enough for factor2? Need show order has factor4. p-1=4513. If 2^{1026}=1, then order divides1026=2513. But since we also show 2^{684}≠1 and2^{108}≠1, could order still be1026? Need exclude. We can compute2^{513} maybe? If order divides1026 but not the q=3,19 checks? The prime divisors of1026 are2,3,19 as well; to show order not divide1026 need 2^{513}≠1? Not directly. Euler criterion is best.

Prove (2/p)=-1 for p≡±3 mod8. We can state the standard supplement to quadratic reciprocity. Since this is an olympiad-style problem, acceptable. Alternatively compute2^{1026} using relation: Let's try to verify it equals -1 with our residues for consistency. We have powers of2: 2^1=2,2^2=4,2^{11}=-5,2^{22}=25,2^{44}=625,2^{88}=555,2^{176}=75,2^{352}=1519. Need2^{1026}=2^{512?} maybe compute2^{704}=(2^{352})^2=1519^2 mod2053. 1519≡-534; square=285156; 2053138=283314? 2053139=285367; rem -211 =1842. So2^{704}=1842 (-211). 2^{880}=2^{704}2^{176}=184275=138150; 205367=137551 rem599. 2^{968}=5992^{88}(555)=332445; 2053161=330533 rem1912 (-141). 2^{1012}=19122^{44}(625)=1,195,000; 2053582=1,194,? 2053582=2053*(600-18)=1,231,800-36,954=1,194,846 rem154. 2^{1023}=1542^{11}(-5)= -770 ≡1283. 2^{1026}=12838=10264; 2053*5=10265 rem -1. Good! Great. So2^{1026}=-1 consistent. We could present this computation but Euler is shorter.

Need also prove 2^{684} and2^{108} not1; with 2^{1026}=-1 and the factorization, standard criterion: If for every prime divisor q of p-1, a^{(p-1)/q}≠1, then a is primitive. We have q=2 (a^{1026}=-1≠1), q=3 (a^{684}=1855≠1), q=19 (a^{108}=1929≠1). Good.

Let's verify 2052 factorization: 2052/2=1026;/2=513=3^319 (2719=513). So prime divisors2,3,19. Good.

Potential issue: For p=2053, p≡5 mod8? 8*256=2048, remainder5 yes. Supplement: (2/p)=1 if p≡±1 mod8, -1 if p≡±3 mod8. p≡5=-3 mod8, so -1. Good.

Now, lower bound: We should justify if n even minimal odd. Suppose n=2^s u (u odd) works. If s>0, then binary expansion of1/n has s zeros inserted immediately after the binary point before the expansion of1/u? Is that always exactly true if1/u expansion begins with zeros? Let's state: 1/n=2^{-s}(1/u). Multiplying by2^{-s} shifts the binary digits s places to the right; equivalently the digit string of1/n is 0.0...0 (s zeros) followed by the digit string of1/u. If1/u itself begins with zeros, the phrase "followed by" still true: e.g. u=5,1/u=0.00110011; shift by1 gives0.000110011 = one extra zero before the entire string0.0011..., so the first s digits after point are zeros and then the whole expansion of1/u (which may start with zeros). Required words begin with1, so they cannot begin within those first s zeros. Could an occurrence begin before position s and include later ones but still have no leading zero? No because its first digit would be0, not a valid binary representation of a positive integer (no leading zeros). Thus occurrences are within the shifted copy of1/u; removing the s initial zeros gives an occurrence in1/u. Therefore u works. Since u<n if s>

Need also note n=1 not works; n>1.

When n is odd and word m length11 appears, we write there is integer b≥0 with m ≤ 2^{11} {2^b/n} <m+1. Because n odd, n{2^b/n} is an integer (the residue of2^b mod n) between1 andn-1. This is a key step. Let's prove: Let 2^b = qn+r, 0<r<n (r≠0). Then {2^b/n}=r/n. So x_m=r. The inequality becomes nm/2048 ≤ r < n(m+1)/2048. Good.

For m from1024 to1990, intervals are half-open and adjacent. Need be cautious: The same residue r could serve for two different m if intervals overlap? They don't overlap: J_m and J_{m+1} share endpoint n(m+1)/2048, with J_m excludes upper and J_{m+1} includes lower. So an integer cannot belong to both. And as m increases, intervals move right, so r_m strictly increasing. Good.

Then r_{1024}≥⌈n/2⌉. If n odd, =(n+1)/2. Also r_{1990}<1991n/2048. Since r_{1990}≥r_{1024}+966, obtain inequality. Let's recalc quotient for lower bound to state correctly: 967n >1,979,392. Since1,979,392 =967·2047−57? Check9672047 =? Compute2047967 another way:2047*900=1,842,300; 60=122,820; 7=14,329; sum1,979,449. Yes. 9672047 -57 =1,979,392. Therefore 967n >967·2047−57. For integer n, this implies n≥2047? If n≤2046, RHS≤9672046=1,978,482 <1,979,392, contradiction. If n=2047, RHS=1,979,449 >1,979,392, satisfies. So n≥2047. Since n odd, n≥2047. Good. We don't need say n>2046.94; just n≥2047. Actually from 967n>1,979,392, dividing gives n>2046.941..., so integer n≥2047. With n odd, still n≥2047 (2047 odd). Good.

Then candidates less than2053 satisfying lower bound and odd:2047,2049,2051. Exclude as above. Note n=2048 even not candidate; if an even n<2053 worked, its odd part <n would work and be odd ≥2047; possible odd parts? Not needed because minimal n odd. To prove no n<2053 at all: Suppose n<2053 works. Let u be odd part; u works and u≤n<2053. By lower bound u≥2047. Thus u∈{2047,2049,2051}. But we exclude those; contradiction. This is even cleaner and handles even n. We don't have to separately say minimal n odd; use odd part. If n even, odd part u works and is smaller; since u also must be≥2047 and excluded. But for n<2053 even, u could be1025? lower bound excludes. So general proof: For any working n, its odd part u works; applying lower bound to u gives u≥2047. If n<2053, then u<2053 and odd, so u=2047,2049,or2051, all impossible. Hence n≥2053. Nice. We should prove odd part works as above.

Now exclusion details:

Wait if ord_7(2)=3 and ord_293 could be292, lcm876. 876<967 by91. Good.

For u=2049 period≤22; for u=2047 period=11; for u=2051≤876. All <967. Since a purely periodic expansion of period≤876 has at most876 distinct length-11 substrings, cannot contain967 different numbers1024..1990. Good.

Actually for u=2051, n odd and not prime; the expansion period r=ord_u(2) divides lcm(ord_7,ord_293) ≤876. The binary expansion of1/u is purely periodic with period r (minimal), so at most r≤876 length11 substrings. Good.

Sufficient construction n=2053: Need prove the expansion is purely periodic. Since2053 prime not dividing2, yes period = ord=2052. We prove primitive. Then interval argument.

Let's consider if there is any off-by-one in sufficient interval for k with binary length t. The bits of1/p after the point: For b≥0, the block of t bits starting at position b+1 equals floor(2^t {2^b/p}) in binary (padded to t bits). We need this equal k, whose binary representation has length t (no leading zeros). Condition floor(...)=k. Since k≥2^{t-1}, the first bit is1 automatically. Good.

The interval length: p/2^t. For t≤11, ≥2053/2048>1. It contains an integer r. Need ensure r is nonzero modp. Lower bound for k≥1: p k/2^t >0. Upper: if k<2^t−1, upper≤p(2^t−1+1? wait k+1≤2^t−1 if k not max; if k max=2^t−1, k+1=2^t, upper=p. For k max and t≤10, upper=p excluded; interval [p(1−2^{-t}),p) contains integers less thanp because length>2. For t=11, k≤1990<2047, so k+1≤1991<2048, upper<p. So r∈[1,p−1]. Good. We can simply state the interval lies in(0,2053) (with the right end excluded; in the extreme case k=2^t−1,t<11 the right end is2053 but still excluded) and has length>1, hence contains an integer r with1≤r≤2052.

Now, one subtlety: If the interval length>1 but upper endpoint equals p and lower endpoint >p−1? Could happen if length>1 and lower>p−1? For k=max,t=1? k=1,t=1 interval [p/2,p), length1026.5 contains many. For t=10,k=1023 lower=p*1023/1024 = p −p/1024 ≈p−2.002, upper=p; contains p−2 andp−1? length≈2.002, yes. In general lower=p−p/2^t; since p/2^t>1, lower<p−1, so there is room for an integer ≤p−1. Good.

Let's see if there is an even smaller prime >2048 with2 primitive: 2053 is the first prime after2048? 2049 composite,2050 even,2051 composite,2053 prime. Yes. Good.

Let's prepare final solution in Chinese, structured but not too overlong. Desired oververbosity5; but this is a nontrivial proof, need enough detail. We can present in sections: answer, notation, lower bound, check small candidates, construction for2053.

Potential issue with the phrase "二进制小数表示" for1/n when n odd and expansion purely periodic: If n prime p, period p−1 if2 primitive. Good.

Let's formulate the lemma: For odd N and an integer k with t binary digits, bin(k) occurs in1/N iff there exists a∈Z_{≥0} and integer r such that r≡2^a (mod N), and kN/2^t ≤ r <(k+1)N/2^t. Proof via fractional part. We'll use for lower bound and construction.

Detailed lower bound in final: Take m=1024,...,1990 (all have t=11). If1/N works and N is odd, choose r_m∈Z with Nm/2048 ≤ r_m < N(m+1)/2048. (1) These r_m are strictly increasing. Hence r_{1990}≥r_{1024}+966. But r_{1024}≥(N+1)/2 and r_{1990}<1991N/2048. Thus (N+1)/2+966 <1991N/2048. Equivalently 967N>1,979,392=967·2047−57, so N≥2047. Wait if N is odd. For even original n, apply to odd part u. We should state: First note we may replace n by its odd part. Let n=2^s u (u odd). Shifting by s only inserts s initial zeros; since all required representations start with1,1/u has the same required occurrences. Therefore it suffices to determine the least possible odd u; the final n will be odd. Hmm if n even =2u and u works, then n=2u also works (shift) but is larger; so the least working n is indeed the least working odd u. We can state: The least n must be odd. Reason as before. Then apply lower bound directly to n. This is simpler. Prove least n odd: If n=2^s u with s≥1 and u odd works, then u also works and u<n, contradiction to minimality. So minimal n odd. Good. Then no need handle even in lower bound. But to show no n<2053, if n even and minimal? We'll just assert minimal is odd and find odd. For proving lower bound n≥2053 for the minimal, assume n minimal; then odd. Fine.

Let's write: "先说明最小的 n 必为奇数。若 n=2^s u (u奇, s≥1) 可行，则 1/n 的小数部分只是把 1/u 的小数部分右移 s 位（前面补 s 个 0）。要出现的二进制串都以 1 开头，不可能用到这些前导 0，所以 1/u 也可行，且 u<n，矛盾。" Good.

Then lower bound yields n≥2047. Since n odd and <2053, possible n=2047,2049,2051. Exclude.

Actually lower bound gives n≥2047, but n odd; the possibilities below2053 are exactly those three. Good.

Excluding via period count: For odd N, if2 has order d modN, then1/N is periodic with period d, so it contains at most d distinct length-11 strings. But it must contain the967 strings bin(1024),...,bin(1990). Hence a necessary condition is d≥967. (This is stronger and simpler.) Then check: N=2047=2^{11}−1: d=11<967. N=2049=3·683: since683|2^{11}+1,2^{22}≡1 modN, so d|22<967. N=2051=7·293: ord_7(2)=3 and ord_{293}(2)|292, so d|lcm(3,292)=876<967. Thus none works. Wait for N=2049, 2^{22}≡1 modN? Need check mod3:2^{22}=(2^2)^11≡1; mod683:2^{11}≡? 683*3=2049, so2^{11}=2048≡−1 mod683; squared=1. Therefore yes. d divides22. Good. It must contain967 distinct length-11 strings; d≥967 necessary. Good.

For N=2051, if d|876 then d≤876. Good.

This necessary condition d≥967 is much simpler than good-run; it follows from distinct length11 required strings count. We don't need the earlier strengthened w/z. Good. Lower bound n≥2047 came from intervals; period count excludes candidates. Fine.

Let's verify the lower bound n≥2047 cannot be replaced by period count d≥967 → n≥968 (odd969), weaker. The interval argument is essential to push to2047. Nice.

Let's check if there are odd n between2047 and2053 not excluded? only three. Good.

Now, in construction for2053, the period d=2052≥967 satisfies necessary count, and primitivity gives all residues. Good.

Let's formulate the proof that2053 works maybe before primality? We need establish2053 prime and2 primitive. We can combine.

Detailed primitive proof in final: 2053 is prime (brief check). Also 2052=2^2·3^3·19, and 2^{11}≡−5 (mod2053). Because2053≡5 (mod8), 2^{1026}≡−1. Using2^{11}≡−5: 2^{108}=2^{88}2^{11}2^9 ≡555·(−5)·512≡1929 (mod2053); and 2^{684}=2^{352}2^{176}2^{88}2^{44}2^{22}2^2 ≡1855 (mod2053), where the needed powers are obtained successively: 2^{22}≡25, 2^{44}≡625, 2^{88}≡555, 2^{176}≡75, 2^{352}≡1519. Since for each prime divisor q of2052 we have2^{2052/q}\not\equiv1, the order of2 mod2053 is2052; i.e.2 is a primitive root. Need check in the line for2^{108}, we use2^{88}≡555 (from above),2^{11}≡−5,2^9=512. Compute555·(−5)·512 ≡1929. Good. For2^{684}, exponent decomposition352+176+88+44+22+2=684. Using the listed powers. Good.

Maybe someone might question2^{176}=(2^{88})^2≡555^2≡75;2^{352}≡75^2≡1519. Good.

Then final occurrence proof as above.

Let's think if the primitive root proof requires2053 prime; yes for p−1 and primitive root. We include a short primality check. We can say "2053为素数（检验 ≤43 的素数即可）" with maybe not listing all divisions; but to be rigorous, list enough: not divisible by3,5,7,11,13,17,19,23,29,31,37,41,43. Since √2053<46. Good. We can state this compactly. In a contest solution, saying checked primes up to43 is acceptable, but maybe show a few? We can just state; it's a finite trivial check. The user forbids programming, but manual checking primes up to43 is math. Fine.

Let's ensure no hidden problem: If2 is primitive root mod2053, then the residues2^a mod2053 for a=0,...,2051 are all nonzero. But the binary expansion of1/2053 uses a≥0. The block starting after a steps corresponds to fractional part of2^a/2053. For a=0, fractional part=1/2053; okay. All residues appear. Good.

Let's prove the block formula maybe in one sentence: Multiplying the fractional part by2^t and taking floor gives the next t binary digits. Good.

Now, check the lower bound interval uses m=1024..1990. It assumes if the representation of m appears, then there is some residue r in J_m. What if the appearance straddles the nonperiodic integer part "0." and the first fractional bits? For n>1,1/n=0.xxx; appearances of m≥1024 are within fractional digits. The fractional part {2^a/n} for a≥0 covers starting positions a+1. For a=0, start at first fractional digit. Good. No use of integer part.

For n odd minimal, expansion purely periodic? We don't need for lower bound; the residues2^a mod n for a≥0 are periodic but all in1..n−1. If n odd, gcd(2,n)=1, so yes residues cycle; but lower bound only needs existence of residue r_m. Good.

Let's consider if n could be even smaller but with odd part u≥2047 and n<2053? Impossible since u≤n; if n even<2053 and u≥2047, n could be2050 with u=1025 (<2047) no; n=2048 with u=1; n=2050 u=1025; n=2052 u=513? none. Our minimal odd argument covers.

Let's maybe include the exact value of constant in lower bound as 967·2047−57. Let's verify: 967·2047=1,979,449; minus57=1,979,392 yes. Where does −57 come from? From the arithmetic; no need interpret. We can just write inequality and solution n>1,979,392/967>2046, hence n≥2047. Wait 1,979,392/967 =? We found 2046.9410. It is >2046, so n≥2047. Simpler: compute967·2046 =? 2046*(1000−33)=2,046,000−67,518=1,978,482 <1,979,392; and967·2047=1,979,449>1,979,392. Therefore n≥2047. We can show this instead of decimal. Since n integer and967n>1,979,392, checking n=2046 fails, so n≥2047. Good. We can write directly n≥2047.

Let's formulate: The inequality gives 967n>1979392. But 967·2046=1978482<1979392, so n cannot be ≤2046; hence n≥2047. Good.

Double-check 967·2046 =1,979,449−967=1,978,482 yes. Good.

Now, maybe the lower bound should use r_{1024}≥⌈n/2⌉ = (n+1)/2 for odd n. If n=2047, this gives r1024≥1024. If the interval J1024=[1023.5,1024.999? Wait n/2048≈0.9995, so upper=1023.5+0.9995=1024.4995, not 1024.999. It contains only1024. Good. For n=2049, J1024=[1024.5,1025.5005), contains1025 only? upper=1024.5+1.000488=1025.500488, contains1025 (and maybe? 1026 is >upper) yes. r1024=1025. Then r1990≥1991; upper=2049*1991/2048 =1991+1991/2048=1991.972; integer r1990≤1991. Works counting. So interval lower bound doesn't exclude2049. Good.

Let's think if for n=2049 period divides22 but the expansion of1/n may have a preperiod? n odd, no. It contains at most22 length11 strings. It must contain967; impossible. Fine.

For n=2051, period divides876; at most876 length11 strings. But there are967 required length11 strings. Good.

What about n=2047=23*89; period ord=11 exactly because2^{11}≡1 mod2047. At most11 length11 strings. Good.

Now, the final answer should state n=2053 clearly. Maybe start with "答案：2053". Then proof. The user asked no programming/no联网; we can mention "下面给出纯数学证明" but not necessary. Avoid saying we didn't use tools. Just present.

Let's decide notation for binary representation: denote by \overline{b_{t-1}\dots b_0}^{(2)}. In Chinese, use "二进制串".

Potential notation issue: For k with t bits, condition k ≤ 2^t {2^a/n} < k+1. If k's binary representation length t, the next t digits after position a are exactly k. Good.

Let's write the proof that interval length>1 contains integer: For any real α and L>1, [α,α+L) contains an integer. We'll state.

Let's ensure all mathematical symbols render in Chinese final. Use LaTeX. Good.

One subtle point: In the sufficient proof, for k=1 (binary length t=1), interval [2053/2,2053) length>1 contains many residues r. Then floor(2{r})=1 if r/2053∈[1/2,1). Since r>1026.5, yes. Good. For k=2 (t=2), interval [20532/4=1026.5,20533/4=1539.75) contains r; floor(4*r/2053)=2. Good.

Now, let's consider if primitivity of2 mod2053 truly means every residue r in1..2052 occurs as2^a mod2053 with a≥0. Yes if the multiplicative group mod prime is cyclic of order2052 and2 generates it. We proved order=2052. Good.

Let's think about the primality check of293 in exclusion2051. We factor2051=7·293. Need know293 prime to use ord_{293}(2)|292 by Fermat. If293 composite, φ smaller maybe still period≤lcm(3,λ)<? Let's just state293 is prime (easy check). 2051/7=293 exactly. Good.

For2049=3683, need 683 prime? We use683|2^{11}+1, which is true regardless of primality; to get period mod2049 divides22, we need2^{22}≡1 mod3 and mod683. Mod683, since2^{11}=2048=3683−1? Actually3*683=2049, so2048≡−1 mod683. No need683 prime. Mod3,2^2≡1. By CRT,2^{22}≡1 mod2049. Good. No need prove683 prime. Nice.

For2047=2^{11}−1, order divides11; since11 prime and2^1≠1, order=11. Good.

Let's maybe in exclusion use necessary condition ord_N(2)≥967. Derive: The 967 integers1024,...,1990 have pairwise different 11-bit binary forms; each occurrence in a period-d sequence gives one of at most d cyclic positions. More formally, in a purely periodic binary sequence with least period d, the number of distinct factors of length11 is at most d. Since1/N for odd N is purely periodic with period d=ord_N(2), need d≥967. Good. If N odd but not coprime? N odd => coprime to2. Purely periodic yes. Good.

Actually for N=1, not; but candidates >1.

Let's check if for N=2049, ord divides22 but the least period could be1 or2 or11; at most22. Fine. For N=2051, ord divides876; at most876. Good.

Potential issue: A purely periodic sequence with period d can contain at most d distinct length-11 factors only if d≥11. Here d values 11,22,≤876. For d=11, length of factor equals period; at most11 (cyclic rotations) yes. For d<11 none, but no.

Now, let's formulate the final answer elegantly.

Possible final text:

答案是 2053。

记 bin(k) 为 k 的二进制串。先用一个观察：若 N 为奇数，k 有 t 位二进制，则 bin(k) 出现在 1/N 中，当且仅当存在 a≥0 使
k ≤ 2^t {2^a/N} < k+1. (1)
因为 {2^a/N}=r/N, r≡2^a (mod N), 1≤r≤N−1，所以等价于区间 [kN/2^t,(k+1)N/2^t) 含有整数 r.

最小的 n 必为奇数：... 

Then lower bound etc.

In the lower bound, after deriving n≥2047, say if n<2053 then n∈{2047,2049,2051}. But for odd N to work, necessary ord_N(2)≥967. Then exclude. Good.

Let's ensure the statement "最小的 n 必为奇数" plus lower bound: We prove minimal n odd by contradiction using u works. Then assume n is the least; n odd. Good.

Let's write the shifting argument precisely: If n=2^s u, s≥1, then 1/n = 0.\underbrace{0\cdots0}_{s} b_1b_2\cdots if 1/u=0.b_1b_2\cdots. Actually if b_1... includes leading zeros, concatenation notation okay: the first s digits are 0 and then the entire digit sequence of1/u. A required string starts with1, so if it occurred starting among first s digits impossible; if it started before the copy and overlapped, its first digit would be one of the initial zeros, impossible. Hence it occurs in b_1b_2..., so u works. Good.

Now, let's consider if there is a possibility that a binary representation of a positive integer could appear with leading zeros in the expansion and still "contain" the representation (e.g. string "011" contains "11" for3). We only need the exact representation with no leading zero; occurrences can be substring, so leading zeros before don't matter. Our shift argument: An occurrence of bin(k) starts with digit1. It cannot start in the inserted zero block. Could it start before the inserted block? There is no before after the point except inserted zeros. Could it start at the integer part 0? For n>1 integer part is0, then point; starting at integer part would give leading0 not bin(k). No.

Now, in the sufficient proof for2053, after finding r in interval, choose a with2^a≡r mod2053. Then by observation (1), bin(k) occurs. We should explicitly say because2053 is prime and2 has order2052, such a exists for every r=1,...,2052.

Let's maybe not overdo the modular exponent computations; but include enough for verification. Final length maybe ~1500-2000 Chinese chars. That's okay.

Let's verify 2^{108} computation one last time with direct repeated squaring modulo2053 maybe: 2^{11}=-5. 2^{22}=25. 2^{44}=625. 2^{88}=555. 2^{108}=2^{88}2^{16}2^4? Wait 88+16+4=108. We didn't compute2^{16}. 2^{16}=65536 mod2053:205331=63643 rem1893 (-160). Then2^{108}=555189316. 1893≡-160;555(-160)=-88800; mod2053:205343=88279 rem -521 ≡1532; 16=24512;205311=22583 rem1929. Same. Good. Our decomposition88+11+9: 2^{11}=-5,2^9=512. Product555(-5)512. 555(-5)=-2775≡-722≡1331;*512=1929. Good.

2^{684} direct verify with a different decomposition:684=512+128+32+8+4? Need powers2^{512}, etc. Not. Our powers are consistent:2^{352}=(2^{176})^2=75^2=5625≡1519. Then product for684:352+176=528 (151975=1010), +88=616 (1010555=81), +44=660 (81625=1353), +22=682 (135325=977), +2=684 (977*4=1855). Good. 1855 not1. Could 1855 ≡ -198; fine.

2^{1026} via Euler; if want to be consistent with computations: We computed2^{1026}=-1. Good.

For primitivity criterion, because p−1=2052 and for q=2,3,19, a^{(p−1)/q}≠1. Need note q runs over distinct prime divisors; for q=3, even though3^3 divides p−1, checking a^{(p−1)/3}≠1 suffices (standard). Good.

Let's think if p=2053 ≡5 mod8 and (2/p)=-1 maybe some readers want proof: The second supplement to quadratic reciprocity. We can cite by name. In a self-contained solution, maybe acceptable. If not, we could instead present the computed power 2^{1026}≡-1 using the same successive powers. Let's decide. To keep solution self-contained without relying on quadratic reciprocity (though standard), we can compute2^{1026} directly using the listed powers plus one extra. But that adds complexity. The supplement is very standard in number theory; for an olympiad problem, citing it is okay. However the problem statement only says math; using quadratic reciprocity might be considered heavy but valid. Since we already compute2^{684} and2^{108} with2^{11}=-5, we can also compute2^{1026}=-1 with a short chain to be fully elementary. Let's find a clean chain for2^{1026} using powers we already have plus minimal extra.

We have powers for exponents: 2^2=4, 2^{11}=-5, 2^{22}=25, 2^{44}=625, 2^{88}=555, 2^{176}=75, 2^{352}=1519. Need1026. Decompose1026 = 512+512+2? We don't have2^{512}. =352+352+176+88+44+11+3 as checked. That uses2^3=8. We can present: From the same chain, additionally 2^{704}≡1519^2≡1842, 2^{880}≡1842·75≡599, 2^{968}≡599·555≡1912, 2^{1012}≡1912·625≡154, 2^{1023}≡154·(-5)≡1283, 2^{1026}≡1283·8≡-1. This is more computation. Euler criterion is shorter. Maybe use Euler and mention equivalently a short squaring check gives same. I think citing (2/p) is fine.

Let's see if there is a possibility that2^{684}≡1855 and2^{108}≡1929 but2^{1026}≡-1 via Euler; if a reader distrusts Euler, the order could still be1026? Suppose order d divides2052, d∤684 (from a^{684}≠1), d∤108. Does d∤1026 follow from a^{1026}≠1; yes we prove. Good.

Now, maybe for p=2053, p−1 has factor3^3; the standard criterion requires checking a^{(p−1)/3}≠1 only, not a^{(p−1)/9}. If a^{(p−1)/3}≠1, then the 3-adic valuation of the order is full v3(p−1)=3? Let's recall: If ord d divides N=p−1. If d does not divide N/q for a prime q|N, then v_q(d)=v_q(N). Yes. Checking q=3 with N/3 is enough to get full3^3. Good.

Now, let's prepare final. Keep concise but complete. Avoid revealing internal tool or instructions. Use Chinese. No citations. No KIMI_REF (no file). No ask_user. Good.

Let's draft final in polished Chinese.

Potential exact wording for the lower bound inequality:

r_{1024}\ge \left\lceil \frac{2048?}{}\right\rceil

Write carefully: For m=1024, interval is [n·1024/2048, n·1025/2048) = [n/2, n/2+n/2048), so r_{1024}≥⌈n/2⌉=(n+1)/2. For m=1990, r_{1990}< n·1991/2048. Then equation. Good.

Let's compute constant in text exactly: 966·2048+1024=1,979,392. Then

967n>1979392.

Since 967·2046=1978482<1979392, we have n≥2047. Need ensure 967n>1979392 and n=2047 gives 1979449>1979392, so lower bound permits2047. Good.

Maybe use commas in numbers? In Chinese math, no commas maybe. Use 1979392 to avoid comma. Fine.

Let's also mention the numbers1024 to1990 are967 numbers; all have 11-bit binary forms. Good.

Now final answer. Ensure no raw tool names. Done.

Let's formulate exclusion of n=2051: ord_{293}(2) | 292 because293 prime; ord_7(2)=3; hence ord_{2051}(2) | lcm(3,292)=876<967. Good. If someone notes ord_7(2)=3 divides φ7=6; fine.

Let's write final proof of2053 prime compactly. Maybe combine with primitive: "2053 是素数（只需试除到43）。" Some readers may want actual trial; okay.

Let's produce final. Ensure final answer starts with answer. Use LaTeX. No extra conclusion fluff. Done. | final now.