# 2011 AMC 12B Problems/Problem 13

## Problem

Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$. What is the sum of the possible values of $w$?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$

## Solution

Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$. This means $a+b+c=9$. $a,b,c$ must be in the set ${1,3,4,5,6}$. The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$. $a+b$, and $b+c$ then must be the remaining two numbers which are $4$ and $6$. The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$.

\begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*}

\begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*}

The sum of the two w's is $15+16=31$ $\boxed{B}$

## Solution 2

Let the four numbers be $z$, $z+a$, $z+b$, and $z+c$. We know that $c$ must be $9$ because that's the greatest difference. So we have $z$, $z+a$, $z+b$, and $z+9$. The 6 possible differences are $a$, $b$, $9$, $b-a$, $9-a$, and $9-b$. We are given that the differences are 1, 3, 4, 5, 6, 9. $a$ and $9-a$ and $b$ and $9-b$ add to 9 which means they have to be 4, 5 and 3,6 or vice versa. Which leaves $1$. That means $b-a$ has to equal $1$. So an and b have to be 3,4, 4,5, or 5,6. For 3,4, we have $z$, $z+3$, $z+4$, and $z+9$. $4z+16=44$. $4z=28$. $z=7$. $w=7+9=16$. Now for 4,5, notice that it doesn't work. The differences are 4, 5, 9, 1, 4, 5. We are missing 6 and 3. For 5,6, it's $z$, $z+5$, $z+6$, and $z+9$. Check that the differences work; they do. We have $4z+20=44$. $4z=24$. $z=6$. $w=6+9=15$. Therefore our answer is $15+16=\boxed{31}$. ~MC413551