AMC10 2024 B

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 $7(m^2+n^2)=7(7^2+6^2)=\boxed{\textbf{(B)}595}$ A similar method is to take $y=1183-26\sqrt{7x}-x^2$, then noting $x=7a^2$ and bashing to find the value of a where x is closest to y.