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

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{{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{120-\sqrt{x}}</math> an integer?
  
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<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ }  11</math>
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<math> \textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) }  11</math>.
  
 
== Solution ==
 
== Solution ==
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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>.  
  
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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>.  
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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>.  
  
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So there are <math>11</math> values for <math>120-\sqrt{x}</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>\boxed{\textbf{(E) }11}</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 E</math>
 
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=9|num-a=11}}
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*[[2006 AMC 12A Problems/Problem 9|Previous Problem]]
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{{AMC10 box|year=2006|ab=A|num-b=9|num-a=11}}
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*[[2006 AMC 12A Problems/Problem 11|Next Problem]]
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 23:23, 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.

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 (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

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