Difference between revisions of "2006 AMC 12A Problems/Problem 10"

Revision as of 19:03, 31 January 2007

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

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

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 $11 \Rightarrow E$

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 == Solution ==
-For <math>\sqrt{120-\sqrt{x}}</math> to be an integer, <math>120-\sqrt{x}</math> must be a perfect square.
+For <math>\sqrt{120-\sqrt{x}}</math> to be an [[integer]], <math>120-\sqrt{x}</math> must be a perfect [[square]].
 Since <math>\sqrt{x}</math> can't be negative, <math>120-\sqrt{x} \leq 120</math>.
-The perfect squares that are less than or equal to <math>120</math> are <math>\{0,1,4,9,16,25,36,49,64,81,100\}</math>.
+The perfect squares that are less than or equal to <math>120</math> are <math>\{0,1,4,9,16,25,36,49,64,81,100\}</math>, so there are <math>11</math> values for <math>120-\sqrt{x}</math>.
-So there are <math>11</math> values for <math>120-\sqrt{x}</math>.
 Since every value of <math>120-\sqrt{x}</math> gives one and only one possible value for <math>x</math>, the number of values of <math>x</math> is <math>11 \Rightarrow E</math>
@@ Line 17: / Line 15: @@
 == See also ==
 * [[2006 AMC 12A Problems]]
-*[[2006 AMC 12A Problems/Problem 9|Previous Problem]]
-*[[2006 AMC 12A Problems/Problem 11|Next Problem]]
+{{AMC box|year=2006|n=12A|num-b=9|num-a=11}}
 [[Category:Introductory Algebra Problems]]

{{{header}}}
Preceded by Problem 9	AMC 12A 2006	Followed by Problem 11