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2000 AMC 10 Problem 20

Problem 20 of 25HarderAlgebraProblem-Solving Techniques

Let A,A, M,M, and CC be nonnegative integers such that A+M+C=10.A + M + C = 10. What is the maximum value of A⋅M⋅C+A⋅M+M⋅C+C⋅A? \begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A? \end{aligned}

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Solution

Notice that A⋅M⋅C+AM+MC+CA=(A+1)(M+1)(C+1)−(A+M+C)−1=(A+1)(M+1)(C+1)−11. \begin{gathered} A \cdot M \cdot C + AM + MC + CA \\ = (A+1)(M+1)(C+1) \\ {}- (A + M + C) - 1 \\ = (A+1)(M+1)(C+1) - 11. \end{gathered} We maximize a product of three positive integers summing to 13.13. The most balanced split is 4,4,5,4, 4, 5, giving 4⋅4⋅5=80.4 \cdot 4 \cdot 5 = 80. The maximum is 80−11=69.80 - 11 = 69. Thus, the correct answer is C.
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