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

Revision as of 22:23, 16 December 2021

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?

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

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 $\boxed{\textbf{(E) }11}$ .

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 1: / Line 1: @@
+{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #10]] and [[2006 AMC 10A Problems/Problem 10|2006 AMC 10A #10]]}}
 == Problem ==
 For how many real values of <math>x</math> is <math>\sqrt{120-\sqrt{x}}</math> an integer?
-<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ }  11</math>
+<math> \textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) }  11</math>.
 == Solution ==
+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>, 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>\boxed{\textbf{(E) }11}</math>.
 == See also ==
-* [[2006 AMC 12A Problems]]
+{{AMC12 box|year=2006|ab=A|num-b=9|num-a=11}}
-*[[2006 AMC 12A Problems/Problem 9|Previous Problem]]
+{{AMC10 box|year=2006|ab=A|num-b=9|num-a=11}}
-*[[2006 AMC 12A Problems/Problem 11|Next Problem]]
+{{MAA Notice}}
 [[Category:Introductory Algebra Problems]]