sod(N) denotes the sum of the base ten digits of N.

N is a 5-digit base ten positive integer divisible by 15, and sod(N) =15

Determine the total count of the values of N for which this is possible.

Well, if sod(N) = 15, then N is automatically divisible by 3.

So, all we need is sod(N) = 15 and N divisible by 5.

In other words, all we need is sod(N) = 15 and N ending in 0 or 5.

This is the same as counting the number of 4 digit numbers with sod = 10 + the number of 4 digit numbers with sod = 15.

But the number of 4 digit numbers with sod = 10 equals the number of 3 digit numbers with SOD between 1 and 10.

And the number of 4 digit numbers with sod = 15 equals the number of 3 digit numbers with SOD between 6 and 15.

So we need

(the number of 3 digit numbers with sod between 1 and 5) +

2*(the number of 3 digit numbers with sod between 6 and 10) +

(the number of 3 digit numbers with sod between 11 and 15)

Nothing especially brilliant occurs to me now.

I counted Charlie's solution to get 35 + 2*184 + 335 = 738