Difference between revisions of "2024 AMC 10B Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | Positive integers <math>x</math> and <math>y</math> satisfy the equation <math>\sqrt{x} + \sqrt{y} = \sqrt{1183}</math>. What is the minimum possible value of <math>x+y</math> | + | Positive integers <math>x</math> and <math>y</math> satisfy the equation <math>\sqrt{x} + \sqrt{y} = \sqrt{1183}</math>. What is the minimum possible value of <math>x+y</math>? |
<math>\textbf{(A) } 585 \qquad\textbf{(B) } 595 \qquad\textbf{(C) } 623 \qquad\textbf{(D) } 700 \qquad\textbf{(E) } 791</math> | <math>\textbf{(A) } 585 \qquad\textbf{(B) } 595 \qquad\textbf{(C) } 623 \qquad\textbf{(D) } 700 \qquad\textbf{(E) } 791</math> |
Latest revision as of 00:53, 20 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 , giving the minimum possible values. The given equation becomes This means that Since this is closest to answer choice , the answer is ~Neoronean ~Tacos_are_yummy_1 (latex)
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.