Question

Find the solution the solutions of the following congruences

a.13x \equiv 20 (mod 33)

b.116x \equiv 41(mod 160)

Answer 1

(a) 13x 20 (mod 33)

So, 5•13x 5•20 (mod 33)

So, 65x 100 (mod 33)

So, - x 1 (mod 33) .

(Since, 65  -1 (mod 33) & 100 1 (mod 33))

So, x - 1 (mod 33)

i.e. x 32 (mod 33) . (Since, 32 - 1 (mod 33)

So, the solution is, x 32 (mod 33)

(b).it has no integer solution, because, gcd(116,160) = 4 but, r doesn't divide 41

So, no solution exists