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:
![\[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hline \mathrm{Zeros}&0&1&2&3&4&6 \end{array}\]](http://latex.artofproblemsolving.com/0/0/4/00411b340b9a09060a6751fbc9f261972418994d.png)
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 
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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