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

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<math>\mathrm{(A)}\ 49 \qquad\mathrm{(B)}\ 59 \qquad\mathrm{(C)}\ 69 \qquad\mathrm{(D)}\ 79 \qquad\mathrm{(E)}\ 89</math>
 
<math>\mathrm{(A)}\ 49 \qquad\mathrm{(B)}\ 59 \qquad\mathrm{(C)}\ 69 \qquad\mathrm{(D)}\ 79 \qquad\mathrm{(E)}\ 89</math>
  
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==Solution==
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==Solution 1==
  
 
The trick is to realize that the sum <math>AMC+AM+MC+CA</math> is similar to the product <math>(A+1)(M+1)(C+1)</math>. If we multiply <math>(A+1)(M+1)(C+1)</math>, we get <cmath>(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.</cmath> We know that <math>A+M+C=10</math>, therefore <math>(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11</math> and <cmath>AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.</cmath> Now consider the maximal value of this expression. Suppose that some two of <math>A</math>, <math>M</math>, and <math>C</math> differ by at least <math>2</math>.  Then this triple <math>(A,M,C)</math> is not optimal. (To see this, WLOG let <math>A\geq C+2.</math> We can then increase the value of <math>(A+1)(M+1)(C+1)</math> by changing <math>A \to A-1</math> and <math>C \to C+1</math>.)
 
The trick is to realize that the sum <math>AMC+AM+MC+CA</math> is similar to the product <math>(A+1)(M+1)(C+1)</math>. If we multiply <math>(A+1)(M+1)(C+1)</math>, we get <cmath>(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.</cmath> We know that <math>A+M+C=10</math>, therefore <math>(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11</math> and <cmath>AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.</cmath> Now consider the maximal value of this expression. Suppose that some two of <math>A</math>, <math>M</math>, and <math>C</math> differ by at least <math>2</math>.  Then this triple <math>(A,M,C)</math> is not optimal. (To see this, WLOG let <math>A\geq C+2.</math> We can then increase the value of <math>(A+1)(M+1)(C+1)</math> by changing <math>A \to A-1</math> and <math>C \to C+1</math>.)
  
 
Therefore the maximum is achieved when <math>(A,M,C)</math> is a rotation of <math>(3,3,4)</math>. The value of <math>(A+1)(M+1)(C+1)</math> in this case is <math>4\cdot 4\cdot 5=80,</math> and thus the maximum of <math>AMC + AM + MC + CA</math> is <math>80-11 = \boxed{\textbf{(C)}\ 69}.</math>
 
Therefore the maximum is achieved when <math>(A,M,C)</math> is a rotation of <math>(3,3,4)</math>. The value of <math>(A+1)(M+1)(C+1)</math> in this case is <math>4\cdot 4\cdot 5=80,</math> and thus the maximum of <math>AMC + AM + MC + CA</math> is <math>80-11 = \boxed{\textbf{(C)}\ 69}.</math>
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==Solution 2==
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We could also do it the more rigorous way, and find out that <math>A = 3</math>, <math>M = 4</math>, and <math>C = 3</math>. This makes sense because to get the largest value of <math>A \cdot M \cdot C</math>, we need to have <math>A</math>, <math>M</math>, and <math>C</math> be as close together as possible. <math>AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69</math>. So the answer is <math>\boxed{\textbf{(C)}\ 69}.</math>
  
 
==See Also==
 
==See Also==

Revision as of 21:05, 6 May 2018

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

We could also do it the more rigorous way, and find out that $A = 3$, $M = 4$, and $C = 3$. This makes sense because to get the largest value of $A \cdot M \cdot C$, we need to have $A$, $M$, and $C$ be as close together as possible. $AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69$. So the answer is $\boxed{\textbf{(C)}\ 69}.$

See Also

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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