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2016 AMC 12B Problem 15

Problem 15 of 25IntermediateAlgebra

All the numbers 2,2, 3,3, 4,4, 5,5, 6,6, 77 are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

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Solution

Pair the opposite faces as (a,b),(c,d),(e,f).(a,b),(c,d),(e,f). Each vertex product uses one face from each pair, so the sum of all eight products factors as (a+b)(c+d)(e+f).(a+b)(c+d)(e+f). The three factors have fixed total 2+3+4+5+6+7=27,2+3+4+5+6+7=27, and a product with fixed sum is largest when the factors are equal, at 99 each. This balance is achievable with (2,7),(3,6),(4,5),(2,7),(3,6),(4,5), giving 9⋅9⋅9=729.9\cdot9\cdot9=729. Thus, the correct answer is D.
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Tagged: factoring · AM-GM Inequality

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