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

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Let <math>A</math>, <math>M</math>, and <math>C</math> be nonnegative integers such that <math>A+M+C=10</math>.  What is the maximum value of <math>A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A</math>?
 
Let <math>A</math>, <math>M</math>, and <math>C</math> be nonnegative integers such that <math>A+M+C=10</math>.  What is the maximum value of <math>A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A</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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<math>\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89</math>
  
 
==Solution 1==
 
==Solution 1==
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==Solution 2 ==
 
==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>
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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, M 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==
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==Solution 3 ==
Lemma: If <math>69</math> is an answer choice on an AMC, then it must be the answer.
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According to our knowledge in AM-GM, the closer numbers are, the value of their product is bigger. Assume that <math>A=B<C</math>, we can get the set <math>A=3,B=3,C=4</math> which the answer is <math>\boxed{\textbf{(C)}\ 69.}</math>
This lemma is very difficult to prove and has been thoroughly analyzed by quantum computing to show it is always true.
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~bluesoul
Since <math>69</math> is an answer choice on this problem, by this lemma, it is the correct answer
 
-Trex
 
  
 
==Video Solution==
 
==Video Solution==
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{{AMC10 box|year=2000|num-b=19|num-a=21}}
 
{{AMC10 box|year=2000|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Revision as of 02:56, 23 June 2022

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

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
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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