AMC10 2006 A
AMC10 2006 A · Q10
AMC10 2006 A · Q10. It mainly tests Exponents & radicals, Vieta / quadratic relationships (basic).
For how many real values of $x$ is $\sqrt{120 - \sqrt{x}}$ an integer?
对于多少个实数 $x$，$\sqrt{120 - \sqrt{x}}$ 是整数？
(A)
3
3
(B)
6
6
(C)
9
9
(D)
10
10
(E)
11
11
Answer
Correct choice: (E)
正确答案：（E）
Solution
(E) Suppose that \(k=\sqrt{120-\sqrt{x}}\) is an integer. Then \(0\le k\le \sqrt{120}\), and because \(k\) is an integer, we have \(0\le k\le 10\). Thus there are 11 possible integer values of \(k\). For each such \(k\), the corresponding value of \(x\) is \((120-k^2)^2\). Because \((120-k^2)^2\) is positive and decreasing for \(0\le k\le 10\), the 11 values of \(x\) are distinct.
（E）设 \(k=\sqrt{120-\sqrt{x}}\) 是整数。则 \(0\le k\le \sqrt{120}\)，且由于 \(k\) 是整数，有 \(0\le k\le 10\)。因此 \(k\) 有 11 个可能的整数取值。对每个这样的 \(k\)，对应的 \(x\) 值为 \((120-k^2)^2\)。因为当 \(0\le k\le 10\) 时，\((120-k^2)^2\) 为正且单调递减，所以这 11 个 \(x\) 值互不相同。
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