Difference between revisions of "2005 AMC 12B Problems/Problem 20"
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Revision as of 14:55, 4 July 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 
.
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 | |
