a) What is the sum of the smallest and second-smallest positive integers satisfying the congruence?

b) Find some integer that is the multiplicative inverse to 160 modulo 1399.

a) 27a = 17 (mod 40)

a = 40n + 11, where n =0, 1, 2, 3, ....etc.

So, the smallest and 2nd smallest positive integers that satisfy the congrurence are:

11 and (40 x 1) + 11 =51

b) 160^(-1) mod 1399 = 1058.

sorry my congruence for a is: 27a = 17 (mod 40)

See a) above: