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2011 AMC 12B Problems/Problem 13: Difference between revisions

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Case 1
<cmath>
<math>
\begin{align*}
(a,b,c)=(3,1,5)\\
z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\
x=w-5\\
4z + a + (a + b) + 9 &= 44\\
y=w-5-1\\
if \hspace{1cm} a &= 3 \\
x=w-5-1-3\\
a + b &= 4\\
w+x+y+z=4w-20=44\\
4z &= 44 - 9 - 3 - 4\\
w=16\\ </math>
z &= 7\\
w &= 16\\
\end{align*}
</cmath>
 
<cmath>
\begin{align*}
if \hspace{1cm} a &= 5\\
a + b &= 6\\
4z &= 44 - 9 - 5 - 6\\
z &= 6\\
w &= 15\\
\end{align*}
</cmath>


Case 2
<math>(a,b,c)=(5,1,3)\\
x=w-3\\
y=w-3-1\\
x=w-3-1-5\\
w+x+y+z=4w-16=44\\
w=15</math>


The sum of the two w's is <math>15+16=31</math> <math>\boxed{B}</math>
The sum of the two w's is <math>15+16=31</math> <math>\boxed{B}</math>

Revision as of 20:23, 2 December 2017

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

\begin{align*} 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 $15+16=31$ $\boxed{B}$

See also

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

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