2015!

How many trailing digits does 2015! have?


The answer is 502.

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1 solution

Caleb Townsend
Mar 4, 2015

To help understand the reasoning behind it, consider: how do trailing zeros exist? When a 2 2 and 5 5 are multiplied, of course. There's always more 2 2 's than 5 5 's, though, so there are as many trailing zeros as there are 5 5 's. Powers of 5 5 contain multiple 5 5 's, so you need to count them for as many 5 5 's as their power: 125 , 125, for example, counts as three 5 5 's. Now here's the general expression for number of trailing zeros: Z ( x ) = n = 1 log 5 ( x ) x 5 n Z(x) = \sum_{n=1}^{\lfloor\log_5(x)\rfloor}\lfloor\frac{x}{5^n}\rfloor Substitute x = 2015 x = 2015 to get Z ( 2015 ) = n = 1 4 2015 5 n = 502 Z(2015) = \sum_{n=1}^{4}\lfloor\frac{2015}{5^n}\rfloor = \boxed{502}

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