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]]}} | {{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{ | + | For how many real values of <math>x</math> is <math>\sqrt{120-\sqrt{x}}</math> an integer? |
| − | <math> \ | + | <math> \textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11</math>. |
== Solution == | == Solution == | ||
| − | For <math>\sqrt{120-\sqrt{x}}</math> to be an | + | 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>. | ||
| Line 12: | Line 12: | ||
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> | + | 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>. |
== See also == | == See also == | ||
Latest revision as of 00:27, 26 September 2024
- 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 
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
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
