How many zeros are there at the end of 100! (the product $1\times 2\times 3\times$ … × 100)?

A trailing zero comes from a pair $2\times 5$, and 2s are far more plentiful than 5s, so just count the factors of 5: ⌊100/5⌋ = 20 multiples of 5, plus ⌊100/25⌋ = 4 numbers that contribute a second 5 (25, 50, 75, 100). Total = 20 + 4 = 24. Stopping at 20 is the usual slip.