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Difference between revisions of "2024 AMC 10B Problems/Problem 13"

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==Solution 1==
 
==Solution 1==
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(Not yet conclusive plz add more stuff)
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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>
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<math>\sqrt{1183} = 13 \sqrt7 = 6\sqrt7 + 7\sqrt7 = \sqrt{252} + \sqrt{343}</math>. <math>252 + 343 = \boxed{\textbf{(B) }595}</math>
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~[[User:Bloggish|Bloggish]]
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2024|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 05:15, 14 November 2024

Problem

Positive integers $x$ and $y$ satisfy the equation $\sqrt{x} + \sqrt{y} = \sqrt{1183}$. What is the minimum possible value of $x+y$.

$\textbf{(A) } 585 \qquad\textbf{(B) } 595 \qquad\textbf{(C) } 623 \qquad\textbf{(D) } 700 \qquad\textbf{(E) } 791$

Solution 1

Note that $\sqrt{1183}=13\sqrt7$. Since $x$ and $y$ are positive integers, and $\sqrt{x}+\sqrt{y}=\sqrt{1183}$, we can represent each value of $\sqrt{x}$ and $\sqrt{y}$ as the product of a positive integer and $\sqrt7$. Let's say that $\sqrt{x}=m\sqrt7$ and $\sqrt{y}=n\sqrt7$, where $m$ and $n$ are positive integers. This implies that $x+y=(\sqrt{x})^2+(\sqrt{y})^2=7m^2+7n^2=7(m^2+n^2)$ and that $m+n=13$. WLOG, assume that ${m}\geq{n}$. It is not hard to see that $x+y$ reaches its minimum when $m^2+n^2$ reaches its minimum. We now apply algebraic manipulation to get that $m^2+n^2=(m+n)^2-2mn$. Since $m+n$ is determined, we now want $mn$ to reach its maximum. Since $m$ and $n$ are positive integers, we can use the AM-GM inequality to get that: $\frac{m+n}{2}\geq{\sqrt{mn}}$. When $mn$ reaches its maximum, $\frac{m+n}{2}={\sqrt{mn}}$. This implies that $m=n=\frac{13}{2}$. However, this is not possible since $m$ and $n$ and integers. Under this constraint, we can see that $mn$ reaches its maximum when $m=7$ and $n=6$. Therefore, the minimum possible value of $x+y$ is $7(m^2+n^2)=7(7^2+6^2)=\boxed{\textbf{(B)}595}$

~Bloggish

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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