2011 AMC 12B Problems/Problem 13
Contents
[hide]Problem
Brian writes down four integers whose sum is . The pairwise positive differences of these numbers are and . What is the sum of the possible values of ?
Solution
Assume that results in the greatest pairwise difference, and thus it is . This means . must be in the set . The only way for 3 numbers in the set to add up to 9 is if they are . , and then must be the remaining two numbers which are and . The ordering of must be either or .
The sum of the two w's is
Solution 2
Let the four numbers be , , , and . We know that must be because that's the greatest difference. So we have , , , and . The 6 possible differences are , , , , , and . We are given that the differences are 1, 3, 4, 5, 6, 9. and and and add to 9 which means they have to be 4, 5 and 3,6 or vice versa. Which leaves . That means has to equal . So an and b have to be 3,4, 4,5, or 5,6. For 3,4, we have , , , and . . . . . Now for 4,5, notice that it doesn't work. The differences are 4, 5, 9, 1, 4, 5. We are missing 6 and 3. For 5,6, it's , , , and . Check that the differences work; they do. We have . . . . Therefore our answer is . ~MC413551
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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All AMC 12 Problems and Solutions |
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