# Difference between revisions of "2016 AMC 8 Problems/Problem 9"

## Problem

What is the sum of the distinct prime integer divisors of $2016$?

$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

## Solutions

### Solution 1

The prime factorization is $2016=2^5\times3^2\times7$. Since the problem is only asking us for the distinct prime factors, we have $2,3,7$. Their desired sum is then $\boxed{\textbf{(B) }12}$.

### Solution 2

We notice that $9 \mid 2016$, since $2+0+1+6 = 9$, and $9 \mid 9$. We can divide $2016$ by $9$ to get $224$. This is divisible by $4$, as $4 \mid 24$. Dividing $224$ by $4$, we have $56$. This is clearly divisible by $7$, leaving $8$. We have $2016 = 9\cdot 4\cdot 7\cdot 8$. We know that $4$ and $8$ are both multiples of $2$, $9$ is $3^2$, and $7$ is prime. This means that the distinct prime factors are $2,3,$ and $7$. Their sum is $\boxed{\textbf{(B) }12}$.