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 $15+16=31$ $\boxed{B}$

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 16
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2011 AMC 12B Problems/Problem 13

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