Difference between revisions of "2006 AMC 12A Problems/Problem 10"
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== Solution == | == Solution == | ||
− | For <math>\sqrt{120-\sqrt{x}}</math> to be an integer, <math>120-\sqrt{x}</math> must be a perfect square. | + | For <math>\sqrt{120-\sqrt{x}}</math> to be an [[integer]], <math>120-\sqrt{x}</math> must be a perfect [[square]]. |
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>. | ||
− | 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> | + | 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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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> | 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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== See also == | == See also == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
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− | + | {{AMC box|year=2006|n=12A|num-b=9|num-a=11}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 18:03, 31 January 2007
Problem
For how many real values of is an integer?
Solution
For to be an integer, must be a perfect square.
Since can't be negative, .
The perfect squares that are less than or equal to are , so there are values for .
Since every value of gives one and only one possible value for , the number of values of is
See also
{{{header}}} | ||
Preceded by Problem 9 |
AMC 12A 2006 |
Followed by Problem 11 |