<s>[INST] Concecutive last zeroes in expansion of $100!$ 
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.
 [/INST] \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.
</s>