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 1
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 that case is finished. 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 has a factor of or one has a factor of . 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 .
Solution 2
If has four divisors, then its divisors would be 1, , and , where and are prime. Therefore, the sum of the divisors of is .
If either or are odd, then or are even. Therefore, and are even, so is a multiple of 4. The only two numbers from the range that are multiples of 4 are 2012 and 2016.
Factoring 2012, we get . To make and even, WLOG, we have and . However, if was 1, then is not prime, so 2012 is not nice.
Factoring 2016, we get . WLOG, we have .
Testing for the lowest , we get and . Therefore, , and , so is nice, with . Therefore, the answer is .
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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