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:
