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 .
Solution 2 (better explanation of 1st)
Trying out the values for a bit leads to us getting with in 1 set for a total of 9 and the other numbers giving a total of 25.
We start off with trying to get a solution, so we only need to find one set with a sum of 4 (the other will automatically have a sum of 4).
We can add 7 to each number. We are now trying to get a set with 32.
Observe that there are 4 odd and 4 even values. To get 32 from 4 numbers , the numbers either have to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26).
If has 2 odd numbers and 2 even numbers we can divide them into two cases: whether has 20 or doesn't have 20.
Case 1: doesn't have 20.
The maximum sum of the 2 even numbers are 2 + 4 = 6 and the maximum sum of the 2 odd numbers are 11 + 13 = 24. This is not enough to reach the 32 we need.
Case 2: has 20.
The minimum sum of the 2 even numbers are 20 + 0 = 20 and the minimum sum of the 2 odd numbers are 5 + 9 = 14. This is already greater than the 32 we need.
We have proven that we cannot partition the 8 numbers into two groups of 4 with the same sum, so the smallest 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 |
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