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

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<math>a+b</math>, and <math>b+c</math> then must be the remaining two numbers which are <math>4</math> and <math>6</math>.
<math>a+b</math>, and <math>b+c</math> then must be the remaining two numbers which are <math>4</math> and <math>6</math>.
The ordering of <math>(a,b,c)</math> must be either <math>(3,1,5)</math> or <math>(5,1,3)</math>.
The ordering of <math>(a,b,c)</math> must be either <math>(3,1,5)</math> or <math>(5,1,3)</math>.


Case 1
Case 1
<math>(a,b,c)=(3,1,5)</math>
<math>
<math>x=w-5</math>
(a,b,c)=(3,1,5)\\
<math>y=w-5-1</math>
x=w-5\\
<math>x=w-5-1-3</math>
y=w-5-1\\
<math>w+x+y+z=4w-20=44</math>
x=w-5-1-3\\
<math>w=16</math>
w+x+y+z=4w-20=44\\
w=16\\ </math>


Case 2
Case 2
<math>(a,b,c)=(5,1,3)</math>
<math>(a,b,c)=(5,1,3)\\
<math>x=w-3</math>
x=w-3\\
<math>y=w-3-1</math>
y=w-3-1\\
<math>x=w-3-1-5</math>
x=w-3-1-5\\
<math>w+x+y+z=4w-16=44</math>
w+x+y+z=4w-16=44\\
<math>w=15</math>
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 20:09, 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 $(5,1,3)$.


Case 1 $(a,b,c)=(3,1,5)\\ x=w-5\\ y=w-5-1\\ x=w-5-1-3\\ w+x+y+z=4w-20=44\\ w=16\\$ (Error compiling LaTeX. Unknown error_msg)

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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