The four factorials 5!, 10!, 15! and 20! are all multiplied together. How many zeros does the resulting product end with?

A trailing zero needs a factor $10=2\times 5$, and inside a factorial the 5s are always the scarce ones, so count those. Using ⌊n/5⌋ + ⌊n/25⌋ + …, the four factorials carry 1, 2, 3 and 4 fives, a total of 1 + 2 + 3 + 4 = 10. The 2s are far more plentiful — 3 + 8 + 11 + 18 = 40 of them — so only 10 pairs can be formed and the product ends in exactly 10 zeros. The trap 40 counts the 2s instead of the 5s. Answering 4 uses the largest factorial alone and forgets the other three, and 9 drops 5! as 'too small to matter', although 5! = 120 already ends in a zero.