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For how many positive integral values of...

For how many positive integral values of n does n! end with precisely 25 zeros?

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`because`Number of zeroes at the end of n!=25 [given]
`impliesE_(5)(x!)=25`
`implies[n/5]+[n/25]+[n/125]+ . . . .=25`
It's easy to see that n=105 is the smallest satisfactory value of n. the next four values of n will also work (i.e., n=106,107,108,109). hence, the answer is 5.
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