2006 AMC 12A Problems/Problem 10

Revision as of 19:31, 5 November 2006 by Xantos C. Guin (talk | contribs) (added solution)

Problem

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

$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ }  11$

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 $11 \Rightarrow E$

See also