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 $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$ ?

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

Solution 1

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 $80-11 = \boxed{\textbf{(C)}\ 69}.$

Solution 2

Notice that if we want to maximize $AMC + AM + MC + AC$ , we want A, M, and C to be as close as possible. For example, if $A = 7, B = 2,$ and $C=1,$ then the expression would have a much smaller value than if we were to substitute $A = 4, B = 5$ , and $C = 1$ . So to make A, B, and C as close together as possible, we divide $\frac{10}{3}$ to get $3$ . Therefore, A must be 3, M must be 3, and C must be 4. $AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69$ . So the answer is $\boxed{\textbf{(C)}\ 69.}$

Solution 3

According to our knowledge in AM-GM, the closer numbers are, the value of their product is bigger. Assume that $A=B<C$ , we can get the set $A=3,B=3,C=4$ which the answer is $\boxed{\textbf{(C)}\ 69.}$ ~bluesoul

Video Solution

https://www.youtube.com/watch?v=Vdou0LpTlzY&t=22s

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 10 Problems and Solutions

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