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?