Difference between revisions of "2024 AMC 10B Problems/Problem 13"
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+ | ==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>? | ||
+ | <math>\textbf{(A) } 585 \qquad\textbf{(B) } 595 \qquad\textbf{(C) } 623 \qquad\textbf{(D) } 700 \qquad\textbf{(E) } 791</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | Note that <math>\sqrt{1183}=13\sqrt7</math>. Since <math>x</math> and <math>y</math> are positive integers, and <math>\sqrt{x}+\sqrt{y}=\sqrt{1183}</math>, we can represent each value of <math>\sqrt{x}</math> and <math>\sqrt{y}</math> as the product of a positive integer and <math>\sqrt7</math>. Let's say that <math>\sqrt{x}=m\sqrt7</math> and <math>\sqrt{y}=n\sqrt7</math>, where <math>m</math> and <math>n</math> are positive integers. This implies that <math>x+y=(\sqrt{x})^2+(\sqrt{y})^2=7m^2+7n^2=7(m^2+n^2)</math> and that <math>m+n=13</math>. WLOG, assume that <math>{m}\geq{n}</math>. It is not hard to see that <math>x+y</math> reaches its minimum when <math>m^2+n^2</math> reaches its minimum. We now apply algebraic manipulation to get that <math>m^2+n^2=(m+n)^2-2mn</math>. Since <math>m+n</math> is determined, we now want <math>mn</math> to reach its maximum. Since <math>m</math> and <math>n</math> are positive integers, we can use the AM-GM inequality to get that: <math>\frac{m+n}{2}\geq{\sqrt{mn}}</math>. When <math>mn</math> reaches its maximum, <math>\frac{m+n}{2}={\sqrt{mn}}</math>. This implies that <math>m=n=\frac{13}{2}</math>. However, this is not possible since <math>m</math> and <math>n</math> and integers. Under this constraint, we can see that <math>mn</math> reaches its maximum when <math>m=7</math> and <math>n=6</math>. Therefore, the minimum possible value of <math>x+y</math> is <math>7(m^2+n^2)=7(7^2+6^2)=\boxed{\textbf{(B)}595}</math> | ||
+ | |||
+ | ~[[User:Bloggish|Bloggish]] | ||
+ | |||
+ | ==Solution 2 (Guessing & Answer Choices)== | ||
+ | Set <math>x=y</math>, giving the minimum possible values. 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 ~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== | ||
+ | {{AMC10 box|year=2024|ab=B|num-b=12|num-a=14}} | ||
+ | {{MAA Notice}} |
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 |
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