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

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2011 AMC 12B Problems/Problem 13

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Problem

Solution

Solution 2

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