2000 AMC 10 Problems/Problem 20

Problem

Let $A$ , $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ . What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$ ?

$\mathrm{(A)}\ 49 \qquad\mathrm{(B)}\ 59 \qquad\mathrm{(C)}\ 69 \qquad\mathrm{(D)}\ 79 \qquad\mathrm{(E)}\ 89$

Solution

The trick is to realize that the sum $AMC+AM+MC+CA$ is similar to the product $(A+1)(M+1)(C+1)$ .

If we multiply $(A+1)(M+1)(C+1)$ , we get $AMC + AM + AC + MC + A + M + C + 1$ .

We know that $(A+M+C)=10$ , therefore $(A+1)(M+1)(C+1) = (AMC + AM + AC + MC) + 11$ .

Therefore the maximum value of $AMC+AM+MC+CA$ is equal to the maximum value of $(A+1)(M+1)(C+1)-11$ . Now we will find this maximum.

Suppose that some two of $A$ , $M$ , and $C$ differ by at least 2. Then this triple $(A,M,C)$ is surely not optimal. Proof: WLOG let $A\geq C+2$ . We can then increase the value of $(A+1)(M+1)(C+1)$ by changing $A\gets A-1$ and $C\gets C+1$ .

Therefore the maximum is achieved in the cases where $(A,M,C)$ is a rotation of $(3,3,4)$ . The value of $(A+1)(M+1)(C+1)$ in this case is $4\cdot 4\cdot 5=80$ . And thus the maximum of $AMC + AM + AC + MC$ is $80-11 = \boxed{69}$ .

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2000 AMC 10 Problems/Problem 20

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