2006 AMC 10A Problem 10

Give the integer a name and undo the square roots one at a time.

1. Say $\sqrt{120 - \sqrt{x}} = k$, where $k$ is a whole number. A square root is never negative, so $k$ is $0$ or more.
2. Square both sides: $120 - \sqrt{x} = k^2$, so
$$\sqrt{x} = 120 - k^2.$$
3. The left side is a square root, so it cannot be negative. That means $k^2$ can be at most $120$. Since $10^2 = 100$ and $11^2 = 121$, $k$ can be $0, 1, 2, \ldots, 10$. That is $11$ values.
4. Each of these $k$ gives exactly one $x$, namely $x = (120 - k^2)^2$, and different $k$ give different $x$. For example, $k = 10$ gives $\sqrt{x} = 20$, so $x = 400$; $k = 0$ gives $\sqrt{x} = 120$, so $x = 14400$.

So there are $11$ values of $x$.

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

When a question asks when an expression is a whole number, name that whole number and work backwards, one layer at a time. At each layer, remember what is allowed (here: square roots are never negative). Don't forget that $0$ is a whole number and a perfect square.

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 $\boxed{\textbf{(E)}\ 11}$.