If a and b are integers and a mod 5 = 2 and b mod 5 = 3, what is ab mod 5?
Am I on the right track? I'm starting with n = dq + r and solving for a & b first:
a = c5 + 2
b = d5 + 3 so,
(c5 + 2)(d5 + 3) mod 5 = solution
(25cd +15c + 10d + 6) mod 5 = solution
we know that 25cd +15c + 10d divides evenlyl by 5 so we just look at the remaining 6, 6 mod 5 = 1
so is ab mod 5 = 1?
