2006 AMC 10A Problem 10

Let $\sqrt{120 - \sqrt{x}} = k$, where $k$ is an integer; since a square root is never negative, $k \ge 0$. Squaring,
$$\sqrt{x} = 120 - k^2.$$
The left side is a square root, so it must be at least $0$: $k^2 \le 120$, which means $k \le 10$ because $11^2 = 121$ is too big. So $k \in \{0, 1, 2, \dots, 10\}$, eleven values.

Each $k$ determines $\sqrt{x} = 120 - k^2 \ge 0$ and hence a single real $x = (120 - k^2)^2$; different $k$ give different $x$. So there are $11$ values of $x$.

The answer is $\boxed{\textbf{(E)}\ 11}$.

"Expression is an integer" problems are usually solved by naming that integer and unwinding the expression one layer at a time, tracking the domain restrictions (here, square roots are nonnegative) at each layer. The count of valid integers is then a count of perfect squares in a range, and $0$ is a perfect square.

• Skipping $k = 0$ ($\sqrt{x} = 120$ is perfectly valid) and answering $10$ (choice (D)).
• Counting negative $k$ as well, or counting only squares of primes or only odd squares, giving one of the small choices.