2024 AMC 10B Problem 13

Factor: $1183 = 7 \cdot 169 = 7 \cdot 13^2$, so $\sqrt{1183} = 13\sqrt{7}$.

Square the equation: $x + y + 2\sqrt{xy} = 1183$. Since $x + y$ is an integer, $\sqrt{xy}$ must be rational, hence an integer, so $xy$ is a perfect square. Then $\sqrt{y} = 13\sqrt7 - \sqrt{x}$ forces $\sqrt{x}$ and $\sqrt{y}$ to be integer multiples of $\sqrt7$: write $\sqrt{x} = m\sqrt7$ and $\sqrt{y} = n\sqrt7$ with positive integers $m + n = 13$.

Then $x + y = 7m^2 + 7n^2 = 7(m^2 + n^2)$. With $m + n$ fixed, $m^2 + n^2 = (m+n)^2 - 2mn = 169 - 2mn$ is smallest when the product $mn$ is largest, i.e. when $m$ and $n$ are as close as possible: $m = 6$, $n = 7$.

$$x + y = 7(36 + 49) = 7 \cdot 85 = 595.$$
(Indeed $x = 252$, $y = 343$ and $\sqrt{252} + \sqrt{343} = 6\sqrt7 + 7\sqrt7 = 13\sqrt7$.)

The answer is $\boxed{\textbf{(B)}\ 595}$.

A sum of square roots of integers can equal a simplified radical only when each root simplifies to the same surd, so the problem is secretly about integers $m + n = 13$. Then "minimize a sum of squares with fixed sum" is the classic balancing principle: values as equal as possible. Recognizing $1183 = 7 \cdot 13^2$ is the only real hurdle.

Ignoring integrality, the minimum would be at $x = y = \frac{1183}{4}$, giving $x + y = 591.5$. The true answer must be at least this and close to it, and $595$ is the only choice near $591.5$; confirm with $m = 6$, $n = 7$.

• Guessing that x and y need not be multiples of 7, or splitting 13 unevenly (like 1 and 12) and reporting a larger sum such as 7 * 145.
• Choosing the answer $585$ because it is below $591.5$ without noticing that no integer solution can beat the real-number minimum.