# 2005 AMC 12B Problems/Problem 20

## 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 $3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}$.

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