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October 26th, 2017, 04:11 AM  #1 
Newbie Joined: Oct 2017 From: Philippines Posts: 2 Thanks: 0  Help with this competition math problem.
For every positive integer n, let s(n) denote the number of terminal zeros in the decimal representation of n!. For example, 10! = 3,628,800 ends in two zeros, so s(10) = 2. How many positive integers less than or equal to 2016 cannot be expressed in the form n + s(n) for some positive integer n?
Last edited by skipjack; October 26th, 2017 at 04:48 AM. 

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