2015 AMC 10B Problems/Problem 23
Contents
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
Solution 1
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 the least are .
.
Solution 3 (Bashing)
We notice that for a to be at the end of a factorial, one multiple of five must be there. Therefore, it is intuitive to start at and work up. If you bash enough you get , , , and . Going any higher will give too many zeros, and then we can stop going higher. .
Solution 4
Let for some natural numbers , such that . Notice that . Thus For smaller , we temporarily let To minimize , we let , then Since , , the only integral value of is , from which we havve .
Now we let and , then Since , .
If , then which is a contradiction.
Thus
Finally, the sum of the four smallest possible and .
~ Nafer
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.