2005 AMC 12B Problems/Problem 20
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
.