Im looking through old exams for a course in Cryptography and have beaten my head against the wall for a long time on one of the questions:(adsbygoogle = window.adsbygoogle || []).push({});

p = 683 is a prime, p-1 = 2*11*31. What is x = 4^11112 mod p?

When i did chinese remainder theorem on primes 2,11,31 i got that x = 16 mod 682, but so far i have not found a way to use this in determining x mod 683..

Also, when computing discrete logs I have found that one goes from mod p to mod (p-1) alot, why is that?

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# A big number modulo a prime

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