Art of Problem Solving
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2006 AMC 10A Problems/Problem 10: Difference between revisions

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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>
Since <math>\sqrt{x}</math> cannot be negative, the outermost [[radicand]] is at most 120. We are interested in the number of integer values that the expression can take, so we count the number of squares less than 120, the greatest of which is <math>10^2=100.</math>


Thus our set of values is  
Thus our set of values is  


<center><math> \{11^2, 10^2, 9^2,\ldots,2^2, 1^2, 0^2\} </math></center>
<center><math> \{10^2, 9^2,\ldots,2^2, 1^2, 0^2\} </math></center>


And our answer is '''11, (E)'''
And our answer is '''11, (E)'''

Revision as of 01:36, 28 February 2007

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 outermost radicand is at most 120. We are interested in the number of integer values that the expression can take, so we count the number of squares less than 120, the greatest of which is $10^2=100.$

Thus our set of values is

$\{10^2, 9^2,\ldots,2^2, 1^2, 0^2\}$

And our answer is 11, (E)

See Also