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2011 AMC 12B Problem 13

Problem 13 of 25IntermediateAlgebraProblem-Solving Techniques

Brian writes down four integers w>x>y>zw \gt x \gt y \gt z whose sum is 44.44. The pairwise positive differences of these numbers are 1,1, 3,3, 4,4, 5,5, 6,6, and 9.9. What is the sum of the possible values for w?w?

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Solution

The largest difference is w−z=9.w-z=9. For either interior number n,n, we have 9=(w−n)+(n−z).9=(w-n)+(n-z). The only pairs among the listed differences that sum to 99 are 3+63+6 and 4+5,4+5, so the remaining difference must be x−y=1.x-y=1. The second largest difference 66 is either w−yw-y or x−z.x-z. If w−y=6,w-y=6, the numbers are {w,w−5,w−6,w−9},\{w,w-5,w-6,w-9\}, so 4w−20=444w-20=44 and w=16.w=16. If x−z=6,x-z=6, the numbers are {w,w−3,w−4,w−9},\{w,w-3,w-4,w-9\}, so 4w−16=444w-16=44 and w=15.w=15. The possible values are 1616 and 15,15, which sum to 31.31. Thus, the correct answer is B.
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