A classic number theory puzzle asks:

How many zeros are at the end of $100!$ when written in base 10?

These trailing zeros come from factors of 10 in the factorial. Each 10 comes from a factor pair of 2 and 5. Since multiples of 2 are more frequent than multiples of 5, we only need to count how many times 5 appears as a factor.

Step-by-Step Count

To count the number of factors of 5 in \(100!\):

\[ \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor + \left\lfloor \frac{100}{125} \right\rfloor + \cdots \]

Calculate:

  • \(\left\lfloor \frac{100}{5} \right\rfloor = 20\)
  • \(\left\lfloor \frac{100}{25} \right\rfloor = 4\)
  • \(\left\lfloor \frac{100}{125} \right\rfloor = 0\)

Total:

\[ 20 + 4 = 24 \]

Final Answer

There are 24 trailing zeros in \(100!\)

This technique generalizes to finding the number of trailing zeros in any factorial.

Reference