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How to count the number of zeros at the end of the number $6^5×2^{12}×5^8×3^8$? To count this I have considered to multiply 5 with even number but still confused to solve it.

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You need a pair of $(2, 5$) to get a $0$ in the end. How many such pairs can you find? What is the limiting factor here? –  Hyperbola Aug 18 '13 at 15:45

1 Answer 1

up vote 4 down vote accepted

HINT:

We have exactly $8$ powers of $5,$

As $5\cdot2=10,$ we need exactly one even number to be paired with each $5$

We clearly have $12>8, 2$s available as multipler

Had the power of $2$ been $<8,$ we could pair each surviving $5$ with one $6$ each

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So we will have 8 zeros at the end ? –  mathphy Aug 18 '13 at 15:47
    
@mathphy, any doubt:) –  lab bhattacharjee Aug 18 '13 at 15:48
    
I'm little skeptic nothing else :D –  mathphy Aug 18 '13 at 15:51
    
@mathphy, what is most important is the method is understandable or not? Right answer will follow the right method –  lab bhattacharjee Aug 18 '13 at 15:54
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@mathphy, the power of $5$ in $200!$ will be $\displaystyle=\sum_{1\le <\infty}\left\lfloor\frac{200}{5^r}\right\rfloor$ Now, $\left\lfloor\frac{200}{5}\right\rfloor=40,$ $\left\lfloor\frac{200}{5^2}\right\rfloor=8,$ $\left\lfloor\frac{200}{5^3}\right\rfloor=1$ and we have much higher power of $2$ available in $200!$ (as in the answer) –  lab bhattacharjee Aug 18 '13 at 17:27

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