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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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\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.
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 $25$ ?
