# Zeros at the End of a Factorial

## Problem

How many trailing zeros does $$n!$$ have?

## Solution

Consider a fairly small factorial, like $$10! = 3,628,800$$, which has $$2$$ trailing zeros. Is there an easy way to see why it has $$2$$ trailing zeros without performing the multiplications directly?

$10! = 10\times9\times8\times7\times6\times5\times4\times3\times2\times1$

Looking at the above, we can see why $$10!$$ has $$2$$ trailing zeros. Obviously, the factor $$10$$ contributes $$1$$ of these. The other trailing zero comes from the product of the factors $$5$$ and $$2$$, which also gives $$10$$.

This idea of counting factors that are multiples of $$5$$ is a good initial approach. Among these, multiples of $$10$$ immediately contribute a trailing zero, while the remaining multiples of $$5$$ contribute a trailing zero when multiplied by an even factor. Since half of the numbers in the factorial’s expansion are even, there is always an even number to pair with a multiple of $$5$$.

Let’s apply this approach to $$15!$$: $15! = 15\times14\times13\times12\times11\times 10\times9\times8\times7\times6\times5\times4\times3\times2\times1$

This factorial contains $$3$$ multiples of $$5$$ ($$5$$, $$10$$, and $$15$$), which gives a total of $$3$$ trailing zeros. Performing the multiplications gives $$15! = 1,307,674,368,000$$, which confirms the solution.

Now, let’s try applying the approach to $$25!$$. We know the expansion of $$25!$$ contains $$5$$ multiples of $$5$$, which are $$5$$, $$10$$, $$15$$, $$20$$, and $$25$$. Thus, applying our method gives $$5$$ trailing zeros. In reality, however, $$25!$$ has $$6$$ trailing zeros: $25!= 15,511,210,043,330,985,984,000,000$

What gives? The issue is that one term in this factorial’s expansion is $$25$$ (or $$5^2$$), which contains two $$5$$s. Thus, it actually contributes $$2$$ trailing zeros to the count instead of just $$1$$. This shows us that we must account for powers of $$5$$ specially. For example, $$5^3=125$$ and $$5^4=625$$ contribute $$3$$ and $$4$$ trailing zeros, respectively.

Fortunately, we can easily represent this process mathematically. If we define $$z_n$$ as the number of trailing zeros for $$n!$$, we can write the formula as follows:

$z_n = \left\lfloor\frac{n}{5}\right\rfloor+ \left\lfloor\frac{n}{5^2}\right\rfloor+ \left\lfloor\frac{n}{5^3}\right\rfloor+ \left\lfloor\frac{n}{5^4}\right\rfloor+ \ldots$

The above formula only needs to be extended to the largest product of $$5$$ that is less than or equal to $$n$$. For example, for $$125!$$ we extend it to the $$3$$3rd term, which gives:

$z_{125} = \left\lfloor\frac{125}{5}\right\rfloor+ \left\lfloor\frac{125}{5^2}\right\rfloor+ \left\lfloor\frac{125}{5^3}\right\rfloor= 25 + 5 + 1 = 31$