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
Analytics only get you so far
