I need help on those problems and hints are also below:

Puzzles 1 Solve the system of congruences below by finding a simultaneous solution: . 2x = 1 mod 5 . 3x = 9 mod 6 . 4x = 1 mod 7 . 5x = 9 mod 11 Puzzles 2 Obtain three consecutive integers, each having a square factor. Puzzles 3 Prove that the congruences x =a modn and x=b mod m admit a simultaneous solution if and only if d = god(m, n) = a -b.Hints 1. You can do this one! Use our theorems. 2. Find an integer a such that Zzla, 32 | (a +1), and 52| (a +2). This is the same as nding a solution to a = 0 mod 4, a = 1 mod 9, a = 2 mod 25 (explain why in your proof). 3. This one is a whole thing. Your proof should explain all of the following claims and more. Make sure you understand the proof you write down. (a) For the forward direction, given x as a solution to the two equations, we have n|x a and m|x I). Then a b = (x b) (x a) is divisible by d. (b) For the backwards direction we have df = a b for some ell E Z. We also know ny = d mod m for some 3: E Z. Multiply by E to get n(y) = d2 mod m = b a mod m. The x = a + (Eyn) will be a solution