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>\ | + | <math>\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89</math> |
− | ==Solution== | + | ==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>.) | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==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, 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 == | ||
+ | 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> | ||
+ | ~bluesoul | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=Vdou0LpTlzY&t=22s | ||
+ | |||
+ | https://www.youtube.com/watch?v=ECzZPuuQMbE ~David | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=19|num-a=21}} | {{AMC10 box|year=2000|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 18:12, 22 April 2023
Problem
Let ,
, and
be nonnegative integers such that
. What is the maximum value of
?
Solution 1
The trick is to realize that the sum is similar to the product
. If we multiply
, we get
We know that
, therefore
and
Now consider the maximal value of this expression. Suppose that some two of
,
, and
differ by at least
. Then this triple
is not optimal. (To see this, WLOG let
We can then increase the value of
by changing
and
.)
Therefore the maximum is achieved when is a rotation of
. The value of
in this case is
and thus the maximum of
is
Solution 2
Notice that if we want to maximize , we want A, M, and C to be as close as possible. For example, if
and
then the expression would have a much smaller value than if we were to substitute
, and
. So to make A, B, and C as close together as possible, we divide
to get
. Therefore, A must be 3, M must be 3, and C must be 4.
. So the answer is
Solution 3
According to our knowledge in AM-GM, the closer numbers are, the value of their product is bigger. Assume that , we can get the set
which the answer is
~bluesoul
Video Solution
https://www.youtube.com/watch?v=Vdou0LpTlzY&t=22s
https://www.youtube.com/watch?v=ECzZPuuQMbE ~David
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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