2009 AMC 12A Problems/Problem 18
- The following problem is from both the 2009 AMC 12A #18 and 2009 AMC 10A #25, so both problems redirect to this page.
Contents
Problem
For , let , where there are zeros between the and the . Let be the number of factors of in the prime factorization of . What is the maximum value of ?
Solution 1
The number can be written as .
For we have . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have .
For we have . For the value in the parentheses is odd, hence .
This leaves the case . We have . The value is obviously even. And as , we have , and therefore . Hence the largest power of that divides is , and this gives us the desired maximum of the function : .
Solution 2
Notice that is a prime factor of an integer if and only if is even. Therefore, given any sufficiently high positive integral value of , dividing by yields a terminal digit of zero, and dividing by 2 again leaves us with where is an odd integer. Observe then that must be the maximum value for because whatever value we choose for , must be less than or equal to .
"Isn't this solution incomplete because we need to show that can be reached?"
An example of 7 being reached is 1000064. 1000064 divided by is 7813.
In fact, 1000064 is the ONLY that satisfies . All others are 6 or lower, because if there are more zeroes, to be divisible by 128 (), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.
Solution 3
As in the first solution, the number can be written as . Factor out of the expression to get .
You can now easily see that is divisible by at least 6 factors of two, from . Any other factors of two will come from the expression .
Make the substitution: . You now have
, so the expression becomes This is valid when .
Obviously, if is negative, the expression will be fractional and not contain any extra factors of two.
If , the value will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.
If , the expression evaluates to . Dividing out twos, you find that this value is only divisible by one factor of 2.
The six factors of two from earlier add to this factor of two to give
Solution written by coolak
Solution 4
Similar to the other solutions, notice that can be written as . Factoring out we see that
Notice that for , will not be an integer, and will "steal" some 's from the . We don't want this to happen, since we want to maximize the exponent of . We start by considering . Then
is an odd number; more specifically, it ends in (all powers of after end in ). Therefore the value in the parentheses will be some large number that ends in . Considering the rules of divisibility, we find that is even, so it is divisible by . Now our exponent of is at . But the divisibility rule for is the last digits of the number must be divisible by . We know the last digits: , which is not divisible by . Therefore is divisible by , but not . Testing more values of , we find that for , the last digit becomes , which means it is not even divisible by . Therefore the highest possible exponent of that we can reach is .
Solution 5 (LTE)
Let . if and if . However, if , then . By LTE, . Since , must equal . So, the answer is
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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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