# 2000 AMC 12 Problems/Problem 12

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

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

[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } [/katex]

## 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 $\boxed{\textbf{(E) }112}.$

## 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 $AMC+AM+AC+MC=64+16+16+16=112$, $\boxed{\text{E}}$.

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 $\boxed{\textbf{(E) }112}$.

## Solution 4 (Semi-rigorous)

Given that $A$, $M$, and $C$ are nonnegative integers, it should be intuitive that maximizing $AMC$ maximizes $AM + MC + CA$. We thus only need to maximize $AMC$. By the AM-GM Inequality, $$\frac{A + M + C}{3} \geq \sqrt[3]{AMC},$$ with equality if and only if $A = M = C$. Note that the maximum of $AMC$ occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing; $A + M + C = 12$ implies that $A = M = C = 4$, so $AMC + AM + MC + CA = 4^3 + 3*4^2 = 112.$ The answer is thus $\boxed{\text{E}}$, as required.

## Solution 5 (Double AM-GM)

We start off the same way as Solution 4, using AM-GM to observe that $AMC \leq 64$. We then observe that

$(A + M + C)^2 = A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144$, since $A + M + C = 12$.

We can use the AM-GM inequality again, this time observing that $\frac{A^2 + M^2 + C^2}{3} \geq \sqrt[3]{{(AMC)}^2}$

Since $AMC \leq 64$, $3 \sqrt[3]{{(AMC)}^2} \leq 48$. We then plug this in to yield

$A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144 \geq 48 + 2(AM + MC + AC)$

Thus, $AM + MC + AC \leq 48$. We now revisit the original equation that we wish to maximize. Since we know $AMC \leq 64$, we now have upper bounds on both of our unruly terms. Plugging both in results in $48 + 64 = \boxed{\textbf{(E) }112}$

## Solution 6 (Optimization)

The largest number for our value would be $A = M = C.$ So $3A = 12$ and $A = M = C = 4.$ $4\times4\times4 + 4\times4 + 4\times4 = 112$ or $\boxed{\textbf{(E) }112}$

~BowenNa