2015 AMC 10B Problems/Problem 23

Problem

Let be a positive integer greater than 4 such that the decimal representation of ends in zeros and the decimal representation of ends in zeros. Let denote the sum of the four least possible values of . What is the sum of the digits of ?

Solution

A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:

We first look at the case when has zero and has zeros. If , has only zeros. But for , has zeros. Thus, and work.

Secondly, we look at the case when has zeros and has zeros. If , has only zeros. But for , has zeros. Thus, the smallest four values of that work are , which sum to . The sum of the digits of is

Solution 2

By Legendre's Formula and the information given, we have that

Trivially, it is obvious that as there is no way that if , would have times as many zeroes as .

First, let's plug in the number We get that , which is obviously not true. Hence,

After several attempts, we realize that the RHS needs to more "extra" zeroes than the LHS. Hence, is greater than a multiple of .

Very quickly, we find that least are .

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See Also

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