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2024 AMC 10B Problems/Problem 13: Difference between revisions

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~[[User:Bloggish|Bloggish]]
~[[User:Bloggish|Bloggish]]
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy


==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 10:03, 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

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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