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Find the smallest positive integer n such that n! ends in 2016 trailing zeros.

 Dec 4, 2019
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8075! = has 2016 trailing zeros.

 

floor(x/5) + floor(x/5^2) + floor(x/5^3) + floor(x/5^4) =2016, solve for x

 

x =8075!.

 Dec 4, 2019
edited by Guest  Dec 4, 2019

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