Difference between revisions of "2024 AMC 10B Problems/Problem 13"
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+ | ==Solution 2 (Guessing & Answer Choices)== | ||
+ | Set <math>x=y</math>. The given equation becomes <cmath>\sqrt{x}=\sqrt{y}=\dfrac{\sqrt{1183}}{2}\implies x=y=\dfrac{1183}{4}.</cmath>This means that <cmath>x+y=\dfrac{1183}{2}=591.5.</cmath>Since this is closest to answer choice <math>\text{(B)}</math>, the answer is <math>\boxed{\textbf{(B) }595}</math> ~Neoronean | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 18:34, 14 November 2024
Contents
[hide]Problem
Positive integers and satisfy the equation . What is the minimum possible value of .
Solution 1
Note that . Since and are positive integers, and , we can represent each value of and as the product of a positive integer and . Let's say that and , where and are positive integers. This implies that and that . WLOG, assume that . It is not hard to see that reaches its minimum when reaches its minimum. We now apply algebraic manipulation to get that . Since is determined, we now want to reach its maximum. Since and are positive integers, we can use the AM-GM inequality to get that: . When reaches its maximum, . This implies that . However, this is not possible since and and integers. Under this constraint, we can see that reaches its maximum when and . Therefore, the minimum possible value of is
Solution 2 (Guessing & Answer Choices)
Set . The given equation becomes This means that Since this is closest to answer choice , the answer is ~Neoronean
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.