Difference between revisions of "2011 AMC 12B Problems/Problem 13"

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m (See also)
 
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The only way for 3 numbers in the set to add up to 9 is if they are <math>1,3,5</math>.  
 
The only way for 3 numbers in the set to add up to 9 is if they are <math>1,3,5</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>.
 
<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>.
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The ordering of <math>(a,b,c)</math> must be either <math>(3,1,5)</math> or <math>(5,1,3)</math>.
  
Case 1
 
<math>(a,b,c)=(3,1,5)</math>
 
<math>x=w-5</math>
 
<math>y=w-5-1</math>
 
<math>x=w-5-1-3</math>
 
<math>w+x+y+z=4w-20=44</math>
 
<math>w=16</math>
 
  
Case 2
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<cmath>
<math>(a,b,c)=(5,1,3)</math>
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\begin{align*}
<math>x=w-3</math>
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z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\
<math>y=w-3-1</math>
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4z + a + (a + b) + 9 &= 44\\
<math>x=w-3-1-5</math>
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\text{if} \hspace{1cm} a &= 3 \\
<math>w+x+y+z=4w-16=44</math>
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a + b &= 4\\
<math>w=15</math>
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4z &= 44 - 9 - 3 - 4\\
 +
z &= 7\\
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w &= 16\\
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\end{align*}
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</cmath>
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 +
<cmath>
 +
\begin{align*}
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\text{if} \hspace{1cm} a &= 5\\
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a + b &= 6\\
 +
4z &= 44 - 9 - 5 - 6\\
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z &= 6\\
 +
w &= 15\\
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\end{align*}
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</cmath>
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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>
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2011|ab=B|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 23:06, 25 December 2018

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

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
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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