2015 AMC 10B Problems/Problem 23

Problem

Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$ ?

$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

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:

$\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hline \mathrm{Zeros}&0&1&2&3&4&6 \end{array}$

We first look at the case when $k!$ has $1$ zero and $(2k)!$ has $3$ zeros. If $k=5,6,7$ , $(2k)!$ has only $2$ zeros. But for $k=8,9$ , $(2k)!$ has $3$ zeros. Thus, $k=8$ and $k=9$ work.

Secondly, we look at the case when $k!$ has $2$ zeros and $(2k)!$ has $6$ zeros. If $k=10,11,12$ , $(2k)!$ has only $4$ zeros. But for $k=13,14$ , $(2k)!$ has $6$ zeros. Thus, the smallest four values of $k$ that work are $k=8,9,13,14$ , which sum to $44$ . The sum of the digits of $44$ is $\boxed{\textbf{(B) 8}}$

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2015 AMC 10B Problems/Problem 23

Problem

Solution