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

(Created page with "==Problem== Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6</m...")
 
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== Solution ==
 
== Solution ==
{{solution}}
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Assume that <math>y-z=a, x-y=b, w-x=c.</math>
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<math>w-z</math> results in the greatest pairwise difference, and thus it is <math>9</math>.
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which means <math>a+b+c=9</math>. <math>a,b,c</math> must be in the set <math>{1,3,4,5,6}</math>.
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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>.
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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>.
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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>.
  
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Case 1
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<math>(a,b,c)=(3,1,5)</math>
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<math>x=w-5</math>
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<math>y=w-5-1</math>
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<math>x=w-5-1-3</math>
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<math>w+x+y+z=4w-20=44</math>
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<math>w=16</math>
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Case 2
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<math>(a,b,c)=(5,1,3)</math>
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<math>x=w-3</math>
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<math>y=w-3-1</math>
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<math>x=w-3-1-5</math>
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<math>w+x+y+z=4w-16=44</math>
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<math>w=15</math>
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The sum of the two w's is <math>15+16=31</math> <math>\boxed{B}</math>
 
== 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}}

Revision as of 22:29, 26 June 2011

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 $9$. which 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 $(a,b,c)=(3,1,5)$ $x=w-5$ $y=w-5-1$ $x=w-5-1-3$ $w+x+y+z=4w-20=44$ $w=16$

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 12 Problems and Solutions
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