Difference between revisions of "2000 AMC 10 Problems/Problem 20"

Revision as of 17:58, 16 May 2020

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 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, B 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

Lemma: If $69$ is an answer choice on an AMC, then it must be the answer. This lemma is very difficult to prove and has been thoroughly analyzed by quantum computing to show it is always true. Since $69$ is an answer choice on this problem, by this lemma, it is the correct answer -Trex

Video Solution

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

@@ Line 13: / Line 13: @@
 ==Solution 2 ==
 Notice that if we want to maximize <math>AMC + AM + MC + AC</math>, we want A, M, and C to be as close as possible. For example, if <math>A = 7, B = 2,</math> and <math>C=1,</math> then the expression would have a much smaller value than if we were to substitute <math>A = 4, B = 5</math>, and <math>C = 1</math>. So to make A, B, and C as close together as possible, we divide <math>\frac{10}{3}</math> to get <math>3</math>. Therefore, A must be 3, B must be 3, and C must be 4.  <math>AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69</math>. So the answer is <math>\boxed{\textbf{(C)}\ 69.}</math>
+==Solution 3==
+Lemma: If <math>69</math> is an answer choice on an AMC, then it must be the answer.
+This lemma is very difficult to prove and has been thoroughly analyzed by quantum computing to show it is always true.
+Since <math>69</math> is an answer choice on this problem, by this lemma, it is the correct answer
+-Trex
 ==Video Solution==

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