Counting Trailing Zeros in a Factorial

A trailing zero at the end of a number comes from a factor of 10, and $10 = 2 \times 5$. In $n!$ (the product of every integer from 1 to $n$), factors of 2 are far more plentiful than factors of 5 — roughly half the numbers up to $n$ are even, while only every fifth number contributes a factor of 5. So the number of trailing zeros in $n!$ is limited entirely by how many times 5 divides into the product, not by how many times 2 does; there will always be more than enough 2s to pair with every 5.

The count of factors of 5 in $n!$ is given by repeatedly dividing by increasing powers of 5 and summing the (rounded-down) results: $\left\lfloor n/5 \right\rfloor + \left\lfloor n/25 \right\rfloor + \left\lfloor n/125 \right\rfloor + \dots$, stopping once the power of 5 exceeds $n$. The extra terms account for numbers like 25 or 125 that contribute more than one factor of 5 each.

For $n = 100$: $\lfloor 100/5 \rfloor = 20$, $\lfloor 100/25 \rfloor = 4$, $\lfloor 100/125 \rfloor = 0$. Summing gives $20 + 4 = 24$, so $100!$ ends in exactly 24 zeros.