For how many positive integers n does n! end in exactly three zeros?

The number of trailing zeros of n! is ⌊n/5⌋ + ⌊n/25⌋ + …, because 5s are scarcer than 2s. This count is 2 for n = 14, becomes 3 at n = 15, and stays 3 for n = 16, 17, 18 and 19, jumping to 4 at n = 20 (which brings a fourth 5). So exactly five values work: 15, 16, 17, 18 and 19. Answering 1 spots only n = 15 and forgets that the count stands still until the next multiple of 5.