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

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Therefore the maximum value of <math>AMC+AM+MC+CA</math> is equal to the maximum value of <math>(A+1)(M+1)(C+1)-11</math>. Now we will find this maximum.
 
Therefore the maximum value of <math>AMC+AM+MC+CA</math> is equal to the maximum value of <math>(A+1)(M+1)(C+1)-11</math>. Now we will find this maximum.
  
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Suppose that some two of <math>A</math>, <math>M</math>, and <math>C</math> differ by at least 2.  
+
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 surely not optimal.
 
Then this triple <math>(A,M,C)</math> is surely not optimal.
 +
 
Proof: 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\gets A-1</math> and <math>C\gets C+1</math>.  
 
Proof: 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\gets A-1</math> and <math>C\gets C+1</math>.  
  
 
Therefore the maximum is achieved in the cases where <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 + AC + MC</math> is <math>80-11 = \boxed{69}</math>.
 
Therefore the maximum is achieved in the cases where <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 + AC + MC</math> is <math>80-11 = \boxed{69}</math>.
 +
 +
<math>\boxed{\text{C}}</math>
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2000|num-b=19|num-a=21}}
 
{{AMC10 box|year=2000|num-b=19|num-a=21}}

Revision as of 11:48, 11 January 2009

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 $80-11 = \boxed{69}$.

$\boxed{\text{C}}$

See Also

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
All AMC 10 Problems and Solutions
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