# 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

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 $A=B=C=4$ Which gives us $$(4+1)(4+1)(4+1)-13=112$$ so the answer is $\boxed{\text{E}}$.