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

m
m (Problem)
 
(8 intermediate revisions by 5 users not shown)
Line 3: Line 3:
 
For how many real values of <math>x</math> is <math>\sqrt{120-\sqrt{x}}</math> an integer?
 
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 ==
 
== 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>.  
 
Since <math>\sqrt{x}</math> can't be negative, <math>120-\sqrt{x} \leq 120</math>.  
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>.  
 
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 \boxed{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>\boxed{\textbf{(E) }11}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 00:27, 26 September 2024

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

Problem

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

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

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

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png