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 $a \cdot b$ , 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 the second case is not possible. We now look at the case of $a \cdot b$ . 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 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 $\frac{2016}{3} - 1$ or $\frac{2010}{3} - 1$ is a prime. We see that in this case none of them work.

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 $2016$ . $2012={4}\cdot{503}$ so either $(a+1)$ or $(b+1)$ both have a factor of $2$ or one has a factor of $4$ since $503$ is prime. If it was the first case, then $a$ or $b$ will equal $1$ . That means that either $(a+1)$ or $(b+1)$ has a factor of $4$ . That means that $a$ or $b$ is $502$ which isn't a prime, so 2012 does not work. $2016 = 4 \cdot 504$ so we have $(503 + 1)(3 + 1)$ . 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is $\boxed{\textbf{(A)}\ 1}$ .

Shortcut

After deducing that $2012$ and case $1$ is impossible, and since there is no option for $0$ , $2016$ is obviously a solution and the answer is $\boxed{\textbf{(A)}\ 1}$ .

-mathboy282

Video Solution by Pi Academy

https://youtu.be/6Pgo0Tvp2Sk?si=2XeBRkt9Xe_QB7v3

~ Pi Academy

Video Solution 2

https://youtu.be/7QNI7VMFkow

~savannahsolver

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=WzSddwBAWtA

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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