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Difference between revisions of "2005 AMC 12B Problems/Problem 20"
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== Solution ==
 
== Solution ==
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The sum of the set is <math>-7-5-3-2+2+4+6+13=8</math>, so if we could have the sum in each set of parenthesis be <math>4</math> then the minimum value would be <math>2(4^2)=32</math>. Considering the set of four terms containing <math>13</math>, this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be <math>13-7-5-3=-2</math>, and with two odd terms then its minimum value is <math>13-7+2-2=6</math>, so we cannot achieve two sums of <math>4</math>. The closest we could have to <math>4</math> and <math>4</math> is <math>3</math> and <math>5</math>, which can be achieved through <math>13-7-5+2</math> and <math>6-3-2+4</math>. So the minimum possible value is <math>3^2+5^2=34\Rightarrow\boxed{C}</math>.
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The sum of the set is <math>-7-5-3-2+2+4+6+13=8</math>, so if we could have the sum in each set of parenthesis be <math>4</math> then the minimum value would be <math>2(4^2)=32</math>. Considering the set of four terms containing <math>13</math>, this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be <math>13-7-5-3=-2</math>, and with two odd terms then its minimum value is <math>13-7+2-2=6</math>, so we cannot achieve two sums of <math>4</math>. The closest we could have to <math>4</math> and <math>4</math> is <math>3</math> and <math>5</math>, which can be achieved through <math>13-7-5+2</math> and <math>6-3-2+4</math>. So the minimum possible value is <math>3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2005|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2005|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:23, 21 December 2020

Problem

Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$

What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$

$\mathrm{(A)}\ 30 \qquad \mathrm{(B)}\ 32 \qquad \mathrm{(C)}\ 34 \qquad \mathrm{(D)}\ 40 \qquad \mathrm{(E)}\ 50$

Solution

The sum of the set is $-7-5-3-2+2+4+6+13=8$, so if we could have the sum in each set of parenthesis be $4$ then the minimum value would be $2(4^2)=32$. Considering the set of four terms containing $13$, this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be $13-7-5-3=-2$, and with two odd terms then its minimum value is $13-7+2-2=6$, so we cannot achieve two sums of $4$. The closest we could have to $4$ and $4$ is $3$ and $5$, which can be achieved through $13-7-5+2$ and $6-3-2+4$. So the minimum possible value is $3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}$.

See also

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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