How many ordered sets of digits (a,b) are there, such that the number 1ab2⎯⎯⎯⎯⎯⎯⎯⎯ is a multiple of 3?
I suppose you mean, 1ab2 in which that line is present over the the set of integers as 1+a+b+2 makes that set of integers.
From the rules of divisibility, we want 1+a+b+2 to be a multiple of 3, hence require a+b to be a multiple of 3.
a and b can be any of the digits from 0 to 9. Out of these, when divided by 3, there are 4 which leave a remainder of 0, 3 which leave a remainder of 1 and 3 which leave a remainder of 2.
If a leaves a remainder of 0, then so must b. This gives us 4×4=16 possibilities.
If a leaves a remainder of 1, then b must leave a remainder of 2. This gives us 3×3=9 possibilities.
If a leaves a remainder of 2, then b must leave a remainder of 1. This gives us 3×3=9 possibilities.
Hence, in total, there are 16+9+9=34 possibilities.