What is the sum of the smallest and second-smallest positive integers $a$ satisfying the congruence$$27a\equiv 17 \pmod{40}~?$$

27*a mod 40 =17, solve for a

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

The smallest "a" =11 and the 2nd smallest "a" =40+11 =51

Thanks for trying but the answer is incorrect

Oh okay the answer is 62

Oh sorry I did not add you answer