2013 AMC 10B Problems/Problem 24
Contents
Problem
A positive integer is nice if there is a positive integer with exactly four positive divisors (including and ) such that the sum of the four divisors is equal to . How many numbers in the set are nice?
Solution
A positive integer with only four positive divisors has its prime factorization in the form of , where and are both prime positive integers or where is a prime. One can easily deduce that none of the numbers are even near a cube so the second case is not possible. We now look at the case of . The four factors of this number would be , , , and . The sum of these would be , which can be factored into the form . Easily we can see that now we can take cases again.
Case 1: Either or is 2.
If this is true then we have to have that one of or is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either or is a prime. We see that in this case none of them work.
Case 2: Both and are odd primes.
This implies that both and are even which implies that in this case the number must be divisible by . This leaves only and . so either or both have a factor of or one has a factor of since is prime. If it was the first case, then or will equal . That means that either or has a factor of . That means that or is which isn't a prime, so 2012 does not work. so we have . 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is .
Shortcut
After deducing that and case is impossible, and since there is no option for , is obviously a solution and the answer is .
-mathboy282
Video Solution
~savannahsolver
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=WzSddwBAWtA
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.