Would the following working help with Question 15?
Let the digits of a 4-digit number n be a, b, c, d.
That is,
Part 1:
If n is divisible by 3, then

for some integer m.
Therefore:
Therefore the sum of the digits (๐ + ๐ + ๐ + ๐) is a multiple of 3.
Part 2:
If the sum of the digits is a multiple of 3, then (๐ + ๐ + ๐ + ๐) = 3๐ for some integer ๐.
Therefore:
Therefore ๐ is divisible by 3. We have now proved the result in both directions, so a 4-digit number is divisible by 3 if and only if the sum of its digits is divisible by 3.