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Problem 7 (When is the order of 2 modulo p equal to 1002). a. Factor 2100 1. Notice that for each prime, p, in this factorization we have 2100 - 1 = 0(mod p), i.e., 2100 = 1 (mod p). This gives us a list of primes that have the potential of 2 having order 100 modulo p (the actual order may be smaller). b. There are exactly three primes p such that 2 has order 100 modulo p. Find them. (i.e., 2100 = 1 (mod p) and 2" 1 (mod p) for any n