How many 0's are at the end of 20!

There is a general formula that can be used. But it is good to get one's hands dirty and compute.

If $20!$ seems dauntingly large, calculate $10!$. You will note it ends with two zeros. Multiplying $10!$ by all the numbers from $11$ to $20$ except $15$ and $20$ will not add to the zeros. Multiplying by $15$ and $20$ will add one zero each.

Remark: Suppose that we want to find the number of terminal zeros in something seriously large, like $2048!$. It is not hard to see that this number is $N$, where $5^N$ is the largest power of $5$ that divides $2048!$. This is because we need a $5$ and a $2$ for every terminal $0$, and the $5$s are the scarcer resource.

To find $N$, it is helpful to think in terms of money. Every number $n$ between $1$ and $2048$ has to pay a $1$ dollar tax for every $5$ "in it." So $45$ has to pay $1$ dollar, but $75$ has to pay $2$ dollars, because $75=5^2\cdot 3$. And a $5$-rich person like $1250$ has to pay $4$ dollars.

Let us gather the tax in stages. First, everybody divisible by $5$ pays a dollar. These are $5$, $10$, $15$ and so on up to $2045$, that is, $5\cdot 1, 5\cdot 2,\dots, 5\cdot 409$. So there are $409$ of them. It is useful to bring in the "floor" or "greatest integer $\le x$ " function, and call the number of dollars gathered in the first stage $\lfloor 2048/5\rfloor$.

But many numbers still owe some tax, namely $25,50,75,\dots,2025$. Get them to pay $1$ dollar each. These are the multiples of $25$, and there are $\lfloor 2048/25\rfloor$ of them.

But $125$, $250$, and so on still owe money. Get them to pay $1$ dollar each. We will gather $\lfloor 2048/125\rfloor$ dollars.

But $625$, $1250$, and $1875$ still owe money. Gather $1$ dollar from each, and we will get $\lfloor 2048/625\rfloor$ dollars.

Now everybody has paid up, and we have gathered a total of $$\lfloor 2048/5\rfloor + \lfloor 2048/25\rfloor +\lfloor 2048/125\rfloor +\lfloor 2048/625\rfloor$$ dollars. That's the number of terminal zeros in $2048!$.

Count up the number of factors of $5$ and the number of factors of $2$ in $20!$. Since we get a zero for every pair of factors $5\cdot 2$, then the minimum of these will answer your question. More simply, $5$ happens less often as a factor (since it's bigger than $2$), so we need only count up the number of $5$'s. In particular, there's one each in $5,10,15,20$, so there are $4$ zeroes at the end.

If the problem had asked about $25!$, then there'd be $6$ zeroes--not $5$--because there are two factors of $5$ in $25$. Similar idea for other numbers.

General formula (for the interested) about the number of zeroes in n! in any base (b). First consider all prime factors of b, then consider the biggest one (p). Then use this formula.

$\lfloor n/p \rfloor$ + $\lfloor n/p^2 \rfloor$ + $\lfloor n/p^3 \rfloor$ + ....

This and using the fact that, the floor becomes zero after some exponent, you can calculate the number of zeroes in any base.

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