2006 AMC 12A Problem 10

Let $k = \sqrt{120 - \sqrt{x}}$ be an integer. Then $k \ge 0$ and $k^2 = 120 - \sqrt{x} \le 120,$ so $0 \le k \le 10.$ Each such $k$ gives $\sqrt{x} = 120 - k^2 \ge 0,$ hence a distinct value $x = (120 - k^2)^2.$ That is $11$ values. Thus, the correct answer is E.