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

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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 E</math>.
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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 ==
 
== See also ==

Revision as of 23:37, 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.

Problem

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

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

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 $11 \Rightarrow \boxed{E}$.

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

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