2013 AMC 10B Problems/Problem 24
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 $a\cdotb$ (Error compiling LaTeX. Unknown error_msg), 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 $a\cdotb$ (Error compiling LaTeX. Unknown error_msg). 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 
.
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 | ||
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