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

Revision as of 22:37, 11 February 2017

The following problem is from both the 2006 AMC 12A #10 and 2006 AMC 10A #10, so both problems redirect to this page.

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 $11 \Rightarrow \boxed{E}$ .

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 12: / Line 12: @@
 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>.
-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>.
+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 \boxed{E}</math>.
 == See also ==