Difference between revisions of "2006 AMC 10A Problems/Problem 10"
(→Solution) |
m (box, minor adjustments) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | For how many real values of <math>\displaystyle x</math> is <math>\sqrt{120-\sqrt{x}}</math> an integer? | + | For how many real values of <math>\displaystyle x</math> is <math>\sqrt{120-\sqrt{x}}</math> an [[integer]]? |
<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 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 | + | Since <math>\sqrt{x}</math> cannot be [[negative]], the outermost [[radicand]] is at most <math>120</math>. We are interested in the number of integer values that the expression can take, so we count the number of squares less than <math>120</math>, the greatest of which is <math>10^2=100</math>. |
Thus our set of values is | Thus our set of values is | ||
| Line 10: | Line 10: | ||
<center><math> \{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 | + | And our answer is <math>11 \Longrightarrow \mathrm{E}</math>. |
| − | == See | + | == See also == |
| − | + | {{AMC10 box|year=2006|ab=A|num-b=9|num-a=11}} | |
| − | |||
| − | |||
| − | |||
| − | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 16:12, 28 February 2007
Problem
For how many real values of 
is 
an integer?
Solution
Since 
cannot be negative, the outermost radicand is at most 
. We are interested in the number of integer values that the expression can take, so we count the number of squares less than , the greatest of which is 
.
Thus our set of values is
And our answer is 
.
See also
| 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 | ||
