# Difference between revisions of "2000 AMC 12 Problems/Problem 12"

## Problem

Let A, M, and C be nonnegative integers such that $\displaystyle A + M + C=12$. What is the maximum value of $A \cdot M \cdot C$+$A \cdot M$+$M \cdot C$+$A\cdot C$?

$\mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 }$

## Solution

$(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + A + M + C + 1$
$(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13$

The term $(A + 1)(M + 1)(C + 1)$ is maximized when A, M, and C are close together, which in this case would be if all of them were 4. Thus,

$125 = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13$
$A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 112 \Rightarrow \mathrm{E}$

 2000 AMC 12 (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 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