Difference between revisions of "2005 AMC 12B Problems/Problem 20"
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== Solution == | == Solution == | ||
| + | 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>. | ||
== See also == | == See also == | ||
* [[2005 AMC 12B Problems]] | * [[2005 AMC 12B Problems]] | ||
Revision as of 14:20, 4 February 2011
Problem
Let 
and 
be distinct elements in the set 
What is the minimum possible value of 
Solution
The sum of the set is 
, so if we could have the sum in each set of parenthesis be 
then the minimum value would be 
. Considering the set of four terms containing 
, 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 
, and with two odd terms then its minimum value is 
, so we cannot achieve two sums of . The closest we could have to and is 
and 
, which can be achieved through 
and 
. So the minimum possible value is 
.
