- #1
GatorPower
- 26
- 0
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:
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?
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?