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I was just trying to work out the exponent for $7$ in the number $343!$. I think the right technique is $$\frac{343}{7}+\frac{343}{7^2}+\frac{343}{7^3}=57.$$ If this is right, can the technique be generalized to $p$ a prime number, $n$ any positive integer, then the exponent of $p$ in $n!$ will be $$\sum_{k=1}^\infty\left\lfloor\frac{n}{p^k}\right\rfloor\quad ?$$ Here, $\lfloor\cdot\rfloor$ denotes the integer less than or equal to $\cdot$ .

Obviously the sum is finite, but I didn't know if it was correct (since its veracity depends on my first solution anyway).

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Yes. Now use this to find the number of trailing zeros :) –  M.B. Oct 14 '13 at 17:55
    
@M.B.: Would I use $p=2$ and $p=5$ in such an instance, then take whichever one is smaller? Also, should I delete the question now? Or should you just write "Yes." as an answer? –  anon271828 Oct 14 '13 at 17:56
    
Yup, I guess you mean the larger of the two? Clearly there are more twos than fives. You should probably give this question a closure, e.g. by replying to it yourself. –  M.B. Oct 14 '13 at 17:58
    
@M.B.: What do you mean by the larger of the two? To count the number of trailing zeros, I need to pair off a $2$ with a $5$. Since I believe your statement about there being more $2$'s than $5$'s is true regardless of the integer $n$, I really only need to count the exponent on $5$, right? This will give me the number of trailing zeros? Thanks. –  anon271828 Oct 14 '13 at 18:01
    
You're right that you count the $5$'s rather than the $2$'s. I guess that either M.B. was just wrong, or they meant that you should take the larger in the sense that $5>2$. A fun way to get an approximate answer is to forget the $\lfloor\rfloor$ signs and then sum the geometric series. So for instance $n!$ has about $n/4$ trailing zeros. –  Oscar Cunningham Oct 14 '13 at 18:09
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Both the solution and its generalization are correct, as @M.B. points out in the comments to the original post. As a corollary, this provides an easy way to count the number of trailing zeros.

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