2015 AMC 10B Problems/Problem 23
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[hide]Problem
Let be a positive integer greater than 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 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 .
We have 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 .
We find that the least four possible are .
.
Solution 3
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 have .
Now we let and , then Since , .
If , then which is a contradiction.
Thus
Finally, the sum of the four smallest possible and .
~ Nafer
Solution 4
We first note that the number of 0's in is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:
The only way we can fulfill the requirements is if and which means that and . The only way this works is if
.
Since we want the smallest values of , we first try it when Thus has 6 zeros, which implies that must have 2. The only way to do this while maintaining our restrictions for is if
So the sum of the four values is so the digit is sum is
-ConfidentKoala4
Solution 5
We will use trial and error to determine the answer to this problem. If , then has zero, and will have zeros. But so does not work. Similarly do not work either. But works because has zero and has zeros. Note that also works because has zero and has zeros. After performing trial and error several times, we find that do not work but do work. Therefore, the four smallest values of are . Therefore adding them together gives and our answer 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 | ||
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