Difference between revisions of "2002 AMC 10B Problems/Problem 16"

(Redirected page to 2002 AMC 12B Problems/Problem 12)
 
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== Problem ==
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#REDIRECT[[2002 AMC 12B Problems/Problem 12]]
 
 
For how many integers <math>n</math> is <math>\frac{n}{20-n}</math> the square of an integer?
 
 
 
<math>\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 10</math>
 
 
 
== Solution ==
 
 
 
For <math>n=20</math> the fraction is undefined, for <math>n>20</math> and <math>n<0</math> it is negative, hence not a square.
 
 
 
This leaves <math>0\leq n < 20</math>.
 
 
 
For <math>n=0</math> the fraction equals <math>0</math>, which is a square.
 
 
 
For <math>1\leq n\leq 9</math> the fraction is strictly between <math>0</math> and <math>1</math>.
 
 
 
For <math>n=10</math> the fraction equals <math>1</math>, which is a square.
 
 
 
The next square is <math>4</math>, and this is achieved for <math>n=16</math>, and the square after that is <math>9</math>, achieved for <math>n=18</math>.
 
 
 
That leaves <math>n=19</math>, for which the fraction is <math>19</math>, which is not a square.
 
 
 
In total, there are <math>\boxed{4}</math> squares among these fractions.
 
 
 
== See Also ==
 
 
 
{{AMC10 box|year=2002|ab=B|num-b=15|num-a=17}}
 

Latest revision as of 18:19, 28 July 2011