# Concecutive last zeroes in expansion of $100!$ [duplicate]

Possible Duplicate:
Highest power of a prime $p$ dividing $N!$

In decimal form, the number $100!$ ends in how many consecutive zeroes. I am thinking of the factorization of $100!$ but I am stuck. I try to count them and since there are 10, 20, 30,..., 100, there are at least 11 zeros. How should I proceed.

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## marked as duplicate by Marvis, Brian M. Scott, Cameron Buie, Old John, PhiraNov 3 '12 at 19:00

And also this math.stackexchange.com/questions/17916/… – user17762 Nov 3 '12 at 18:16

\begin{align} 2\text{ goes into }100 & & 50\text{ times} \\ 2^2\text{ goes into }100 & & 25\text{ times} \\ 2^3\text{ goes into }100 & & 12\text{ times} \\ 2^4\text{ goes into }100 & & 6\text{ times} \\ 2^5\text{ goes into }100 & & 2\text{ times} \\ 2^6\text{ goes into }100 & & 1\text{ time} \\ \end{align} $$50+25+12+6+2+1 = 96.$$

Thus $2^{96}$ divides $100!$ and $2^{97}$ does not.

\begin{align} 5\text{ goes into }100 & & 20\text{ times} \\ 5^2\text{ goes into }100 & & 4\text{ times} \end{align}

Thus $5^{24}$ divides $100!$ and $5^{25}$ does not.

$$\min\{96,24\}=24.$$

So $(2\cdot5)^{24}$ divides $100!$ and $(2\cdot5)^{25}$ does not.

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You want to know the highest power of $10$ that goes into $100!$, so figure out the highest power of $5$ that goes into $100!$. To do this, think:

1) How many integers between $1$ and $100$ are multiples of $5$ ?

2) How many integers between $1$ and $100$ are multiples of $25$ ?

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