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

(New page: 16. 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\q...)
 
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16. For how many integers <math>n</math> is <math>\frac{n}{20-n}</math> the square of an integer?
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== Problem ==
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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>
 
<math>\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 10</math>
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== Solution ==
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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.
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This leaves <math>0\leq n < 20</math>.
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For <math>n=0</math> the fraction equals <math>0</math>, which is a square.
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For <math>1\leq n\leq 9</math> the fraction is strictly between <math>0</math> and <math>1</math>.
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For <math>n=10</math> the fraction equals <math>1</math>, which is a square.
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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>.
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That leaves <math>n=19</math>, for which the fraction is <math>19</math>, which is not a square.
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In total, there are <math>\boxed{4}</math> squares among these fractions.
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== See Also ==
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{{AMC10 box|year=2002|ab=B|num-b=15|num-a=17}}

Revision as of 07:40, 2 February 2009

Problem

For how many integers $n$ is $\frac{n}{20-n}$ the square of an integer?

$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 10$

Solution

For $n=20$ the fraction is undefined, for $n>20$ and $n<0$ it is negative, hence not a square.

This leaves $0\leq n < 20$.

For $n=0$ the fraction equals $0$, which is a square.

For $1\leq n\leq 9$ the fraction is strictly between $0$ and $1$.

For $n=10$ the fraction equals $1$, which is a square.

The next square is $4$, and this is achieved for $n=16$, and the square after that is $9$, achieved for $n=18$.

That leaves $n=19$, for which the fraction is $19$, which is not a square.

In total, there are $\boxed{4}$ squares among these fractions.

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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