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

Revision as of 13:57, 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

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

@@ Line 14: / Line 14: @@
 ~ 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}}

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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