2024 AMC 10B Problem 13

1. Simplify $\sqrt{1183}$. Try small primes: $1183 = 7 \cdot 169 = 7 \cdot 13^2$. So $\sqrt{1183} = 13\sqrt7$.
2. Why $x$ must be $7$ times a square. Move $\sqrt{x}$ over: $\sqrt{y} = 13\sqrt7 - \sqrt{x}$. Square both sides:
$$y = 1183 - 26\sqrt{7x} + x.$$
Here $x$ and $y$ are whole numbers, so $26\sqrt{7x}$ must be a whole number too. The square root of a whole number is either a whole number or not even a fraction, so $\sqrt{7x}$ is a whole number and $7x$ is a perfect square. That means $x$ is $7$ times a perfect square: $x = 7m^2$, so $\sqrt{x} = m\sqrt7$.
3. The same goes for $y$: $\sqrt{y} = n\sqrt7$ for a positive whole number $n$. Then $m\sqrt7 + n\sqrt7 = 13\sqrt7$, so
$$m + n = 13.$$
4. What to make small. $x + y = 7m^2 + 7n^2 = 7(m^2 + n^2)$. So we need $m^2 + n^2$ as small as possible with $m + n = 13$. Make a table:

5. The smallest is $85$, when $m$ and $n$ are as close as possible. So
$$x + y = 7 \cdot 85 = 595.$$
6. Check: $x = 7 \cdot 36 = 252$ and $y = 7 \cdot 49 = 343$, and $\sqrt{252} + \sqrt{343} = 6\sqrt7 + 7\sqrt7 = 13\sqrt7$.

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

Two square roots of whole numbers can add up to $13\sqrt7$ only if each one is a whole number times $\sqrt7$. That turns the problem into splitting $13$ into two whole numbers. When the sum of two numbers is fixed, the sum of their squares is smallest when the numbers are as equal as possible, as the table shows.

Estimate first. If $x$ and $y$ did not have to be whole numbers, the smallest total would come from $x = y$: then $\sqrt{x} = \frac12\sqrt{1183}$, so $x = \frac{1183}{4}$ and $x + y = \frac{1183}{2} = 591.5$. The real answer must be at least $591.5$ and close to it. That rules out $585$ and makes $595$ the natural candidate, which the check $x = 252$, $y = 343$ confirms.