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How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?

How many zeroes does $2012!$ end with?

My idea is:

402 zeroes come from $2\times 5$, 80 from $2\times 25$, 16 from $2\times 125$ and 3 from $2\times 625$

How can we "show" that this is true?

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marked as duplicate by Marvis, Qiaochu Yuan May 5 '12 at 20:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Asked here the general case for $n!$(Derive a formula to find the number of trailing zeroes in $n!$) –  Américo Tavares Mar 13 '12 at 16:34
You know that you can accept one of the answers by clicking on the tick mark (right?) Doesn't have to be necessarily mine –  Kirthi Raman Mar 14 '12 at 14:30
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3 Answers

up vote 12 down vote accepted

The correct answer is 501.

In order to find the number of zeros is same as finding the number of factors of powers of $5$. There are more factors of powers of $2$ than the factors of powers of $5$.

For instance $10! = 3628800 = \hspace{3pt}2^8 \hspace{3pt} 3^4 \hspace{3pt}5^2\hspace{3pt} 7$

$$\left \lfloor \frac{n}{p} \right \rfloor +\left \lfloor \frac{n}{p^2} \right \rfloor +\left \lfloor \frac{n}{p^3}\right \rfloor + \cdots \left \lfloor \frac{n}{p^{k-1}} \right \rfloor$$

where $\left \lfloor \frac{n}{p^k} \right \rfloor=0$. In this case $k=5$ because $\left \lfloor \frac{2012}{5^5} \right \rfloor=0$

$$\left \lfloor \frac{2012}{5} \right \rfloor =402, \hspace{6pt} \left \lfloor \frac{2012}{5^2} \right \rfloor = \left \lfloor \frac{402}{5} \right \rfloor =80$$

$$\left \lfloor \frac{2012}{5^3} \right \rfloor = \left \lfloor \frac{80}{5} \right \rfloor =16, \hspace{6pt} \left \lfloor \frac{2012}{5^4} \right \rfloor = \left \lfloor \frac{16}{5} \right \rfloor =3$$

$$\left \lfloor \frac{2012}{5} \right \rfloor+ \left \lfloor \frac{2012}{5^2}\right \rfloor +\left \lfloor \frac{2012}{5^3}\right \rfloor +\left \lfloor \frac{2012}{5^4} \right \rfloor = 402+80+16+3=501$$

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+1 for being reader friendly, Oh, well, we don't expect people to follow suit. But, may I ask you that you'll replace this with a MathJaX version when it recovers? –  user21436 Mar 13 '12 at 13:27
Oh yes, I also would prefer that. –  Kirthi Raman Mar 13 '12 at 13:32
Also, I'd recommend \left \lfloor \frac{...}{...} \right \rfloor for writing the floor of a fraction in LaTeX. –  Ilmari Karonen Mar 13 '12 at 13:55
Thanks Kirthi Raman! –  VVV Mar 14 '12 at 16:15
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As you have noticed, zeros come only from the product of an even integer with a multiple of 5. Since there are less multiples of 5 than multiples of 2, you only have to consider multiples of 5. Now multiples of powers of 5 contribute more than one zero. So the number of zeros in $n!$ is $$ \def\F#1{\left\lfloor \frac{n}{#1}\right\rfloor} \F{5}+ \F{5^2} + \F{5^3} + \cdots $$

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Please check "there are more multiples of 5 than multiples of 2". I think there is a typo: "there are less multiples of 5 than multiples of 2" –  Américo Tavares Mar 13 '12 at 16:39
@AméricoTavares, fixed, thanks! –  lhf Mar 13 '12 at 18:21
Thanks, lhf... ! –  VVV Mar 14 '12 at 16:14
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The power of any prime $p$ in $n!$ is given by

$[\frac{n}{p} ] + [\frac{n}{p^2}] + [\frac{n}{p^3}].......$

The number of zeros in $2012!$is the number of times 5 occurs in its prime factorization

$[\frac{2012}{5} ] + [\frac{2012}{25}] + [\frac{2012}{125}] + [\frac{2012}{625}] + [\frac{2012}{3125}] $= $402 + 16+ 80 + 3 = 501$

hence $2012! $ has $501$ zeros.

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Thanks 5ToM...! –  VVV Mar 14 '12 at 16:14
Your most welcome tripleV! –  Tomarinator Mar 14 '12 at 16:41
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