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

(Solution)
(Problem)
Line 1: Line 1:
 
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #10]] and [[2006 AMC 10A Problems/Problem 10|2006 AMC 10A #10]]}}
 
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #10]] and [[2006 AMC 10A Problems/Problem 10|2006 AMC 10A #10]]}}
 
== Problem ==
 
== Problem ==
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{144-\cbrt{x}}</math> an integer?
  
 
<math> \textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) }  11</math>.
 
<math> \textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) }  11</math>.

Revision as of 19:50, 21 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 $\sqrt{144-\cbrt{x}}$ (Error compiling LaTeX. Unknown error_msg) 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