AMC12 2006 A

AMC12 2006 A · Q10

AMC12 2006 A · Q10. It mainly tests Exponents & radicals, Functions basics.

For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?

有多少个实数 $x$ 使得 $\sqrt{120-\sqrt{x}}$ 是整数？

(A) 3 3

(B) 6 6

(D) 10 10

(E) 11 11

Answer

Correct choice: (E)

正确答案：（E）

Solution

For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square. Since $\sqrt{x}$ can't be negative, $120-\sqrt{x} \leq 120$. The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$, so there are $11$ values for $120-\sqrt{x}$. Since every value of $120-\sqrt{x}$ gives one and only one possible value for $x$, the number of values of $x$ is $\boxed{\textbf{(E) }11}$.

要使 $\sqrt{120-\sqrt{x}}$ 为整数，$120-\sqrt{x}$ 必须是一个完全平方数。由于 $\sqrt{x}$ 不能为负，故 $120-\sqrt{x} \leq 120$。不超过 $120$ 的完全平方数为 $\{0,1,4,9,16,25,36,49,64,81,100\}$，因此 $120-\sqrt{x}$ 有 $11$ 种取值。每一个 $120-\sqrt{x}$ 的取值都对应唯一的 $x$，因此 $x$ 的取值个数为 $\boxed{\textbf{(E) }11}$。

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