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

 
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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>
 
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math>
 
== Solution ==
 
== Solution ==
 +
Since <math>\sqrt{x}</math> cannot be negative, the only integers we get can from our expression are square roots less than 120. The highest is
 +
 +
<math>11^2=121</math>
 +
Thus our set of values is
 +
 +
{<math>11^2, 10^2, 9^2,....2^2, 1^2, 0^2</math>}
 +
 +
And our answer is '''11, (E)'''
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]

Revision as of 16:01, 3 July 2006

Problem

For how many real values of $\displaystyle x$ 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

Since $\sqrt{x}$ cannot be negative, the only integers we get can from our expression are square roots less than 120. The highest is

$11^2=121$ Thus our set of values is

{$11^2, 10^2, 9^2,....2^2, 1^2, 0^2$}

And our answer is 11, (E)

See Also

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