I need help on those puzzle and hints also given below:

You should be extra careful to justify why every line in your proof is true. Puzzles 1 Let n = pl 19; . . . p\" be composite where the p,- are ail distinct primes. Show that a = b mod n if and only if for every prime divisor p of n, we have a = b mod p. Puzzlw 2 Prove that #5,"), and 11 do not divide it, then 385\":60 1. P111213 3 Let p, g be distinct odd primes. Suppose that neither p nor :9 divide a. Prove that a'" = 1 mod pq where m = W. Puzzlw 4 Let p be an odd prime. Show that (p_1):=(p1) mod1+2+3+-~+(p-1)- Hints 1. You can do this one! 2. Note that 385 = 5 . 7 . 11. Use the above puzzle and Fermat's little theorem. 3. Show that a" = 1 mod p and am = 1 mod q to conclude as before that a" = 1 mod pq. For each of those congruences use Fermat's theorem. 4. Show that (p - 1)! = p-1 mod p and (p-1)! = (p-1) mod ,. This means that (p - 1)! - (p -1) is a solution to the system x=0 mod p x =0 mod (p -1)/2. Recall that the Chinese Remainder Theorem says that any two solutions to the above system are the same modulo p(p - 1)/2. We know (p -1)! - (p-1) is a solution. Can you think of another solution to the system above