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

## Problem

Let $A, M,$ and $C$ be nonnegative integers such that $A + M + C=12$. 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 1

It is not hard to see that $$(A+1)(M+1)(C+1)=$$ $$AMC+AM+AC+MC+A+M+C+1$$ Since $A+M+C=12$, we can rewrite this as $$(A+1)(M+1)(C+1)=$$ $$AMC+AM+AC+MC+13$$ So we wish to maximize $$(A+1)(M+1)(C+1)-13$$ Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$, we set $A=M=C=4$ Which gives us $$(4+1)(4+1)(4+1)-13=112$$ so the answer is $\boxed{\text{E}}$. I wish you understand this problem and can use it in other problems.

## Solution 2 (Nonrigorous)

If you know that to maximize your result you $\textit{usually}$ have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make $A,M$ and $C$ as close as possible. In this case, they would all be equal to $4$, so $AMC+AM+AC+MC=64+16+16+16=112$, giving you the answer of $\boxed{\text{E}}$.

## Solution 3

Assume $A$, $M$, and $C$ are equal to $4$. Since the resulting value of $AMC+AM+AC+MC$ will be $112$ and this is the largest answer choice, our answer is $\boxed{\textbf{(E) }112}$.

## See also

 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

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