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 $(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.$ We know that $A+M+C=10$ , therefore $(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11$ and $AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.$ Now consider the maximal value of this expression. Suppose that some two of $A$ , $M$ , and $C$ differ by at least $2$ . Then this triple $(A,M,C)$ is not optimal. (To see this, WLOG let $A\geq C+2.$ We can then increase the value of $(A+1)(M+1)(C+1)$ by changing $A \to A-1$ and $C \to C+1$ .)

Therefore the maximum is achieved when $(A,M,C)$ 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 $AMC + AM + MC + CA$ is $80-11 = \boxed{\textbf{(C)}\ 69}.$

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

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