2013 AMC 10B Problems/Problem 24

Problem

A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$ ) such that the sum of the four divisors is equal to $n$ . How many numbers in the set $\{ 2010,2011,2012,\dotsc,2019 \}$ are nice?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$

Solution

A positive integer with only four positive divisors has its prime factorization in the form of $a*b$ , where $a$ and $b$ are both prime positive integers or c^3 where c 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 $a*b$ . The four factors of this number would be $1$ , $a$ , $b$ , and $ab$ . The sum of these would be $ab+a+b+1$ , which can be factored into the form $(a+1)(b+1)$ . Easily we can see that Now we can take cases again.

Case 1: Either $a$ or $b$ is 2.

If this is true then we have to have that one of $(a+1)$ or $(b+1)$ is even and that one is 3. So we have that in this case the only numbers that work are odd multiples of 3 which are 2010 and 2016. So we just have to check if either $\frac{2016}{3} - 1$ or $\frac{2010}{3} - 1$ is a prime. We see that $2016$ is the only one that works in this case.

Case 2: Both $a$ and $b$ are odd primes.

This implies that both $(a+1)$ and $(b+1)$ are even which implies that in this case the number must be divisible by $4$ . This leaves only $2012$ and $2016$ . We know $2016$ works so it suffices to check whether $2012$ works. $2012=4*503$ so we have that a factor of $2$ must go to both $(a+1)$ and $(b+1)$ . So we have that $(a+1)$ and $(b+1)$ equal the numbers $(2+503)(2+1)$ , but this contradicts our assumption for the case. Thus the answer is $\boxed{\textbf{(A)}\ 1}$ as $2016$ is the only solution.

Note: 2016/3-1=672-1=671=61*11, so why did we get the right answer?

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2013 AMC 10B Problems/Problem 24

Problem

Solution