Difference between revisions of "2000 AMC 10 Problems/Problem 20"
(→Solution 3) |
(→Video Solution) |
||
(7 intermediate revisions by 5 users not shown) | |||
Line 3: | Line 3: | ||
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 1== | ==Solution 1== | ||
Line 12: | Line 12: | ||
==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, | + | 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== | ==Video Solution== | ||
https://www.youtube.com/watch?v=Vdou0LpTlzY&t=22s | https://www.youtube.com/watch?v=Vdou0LpTlzY&t=22s | ||
+ | |||
+ | https://www.youtube.com/watch?v=ECzZPuuQMbE ~David | ||
==See Also== | ==See Also== | ||
Line 21: | Line 27: | ||
{{AMC10 box|year=2000|num-b=19|num-a=21}} | {{AMC10 box|year=2000|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 17:12, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.