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Difference between revisions of "2002 AMC 12B Problems/Problem 12"

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\qquad\mathrm{(E)}\ 10</math>
 
\qquad\mathrm{(E)}\ 10</math>
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
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Let <math>x^2 = \frac{n}{20-n} </math>, with <math>x \ge 0</math> (note that the solutions <math>x < 0</math> do not give any additional solutions for <math>n</math>). Then rewriting, <math>n = \frac{20x^2}{x^2 + 1}</math>. Since <math>\text{gcd}(x^2, x^2 + 1) = 1</math>, it follows that <math>x^2 + 1</math> divides <math>20</math>. Listing the factors of <math>20</math>, we find that <math>x = 0, 1, 2 , 3</math> are the only <math>4 \Rightarrow \mathrm{(D)}</math> solutions (respectively yielding <math>n = 0, 10, 16, 18</math>).
 
Let <math>x^2 = \frac{n}{20-n} </math>, with <math>x \ge 0</math> (note that the solutions <math>x < 0</math> do not give any additional solutions for <math>n</math>). Then rewriting, <math>n = \frac{20x^2}{x^2 + 1}</math>. Since <math>\text{gcd}(x^2, x^2 + 1) = 1</math>, it follows that <math>x^2 + 1</math> divides <math>20</math>. Listing the factors of <math>20</math>, we find that <math>x = 0, 1, 2 , 3</math> are the only <math>4 \Rightarrow \mathrm{(D)}</math> solutions (respectively yielding <math>n = 0, 10, 16, 18</math>).
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=== Solution 2 ===
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For <math>n<0</math> and <math>n>20</math> the fraction is negative, for <math>n=20</math> it is not defined, and for <math>n\in\{1,\dots,9\}</math> it is between 0 and 1.
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Thus we only need to examine <math>n=0</math> and <math>n\in\{10,\dots,19\}</math>.
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For <math>n=0</math> and <math>n=10</math> we obviously get the squares <math>0</math> and <math>1</math> respectively.
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For prime <math>n</math> the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
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This leaves <math>n\in\{12,14,15,16,18\}</math>, and a quick substitution shows that out of these only <math>n=16</math> and <math>n=18</math> yield a square.
  
 
== See also ==
 
== See also ==

Revision as of 15:29, 6 January 2009

Problem

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

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

Solution

Solution 1

Let $x^2 = \frac{n}{20-n}$, with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$. Since $\text{gcd}(x^2, x^2 + 1) = 1$, it follows that $x^2 + 1$ divides $20$. Listing the factors of $20$, we find that $x = 0, 1, 2 , 3$ are the only $4 \Rightarrow \mathrm{(D)}$ solutions (respectively yielding $n = 0, 10, 16, 18$).

Solution 2

For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $n\in\{1,\dots,9\}$ it is between 0 and 1.

Thus we only need to examine $n=0$ and $n\in\{10,\dots,19\}$.

For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.

For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.

This leaves $n\in\{12,14,15,16,18\}$, and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square.

See also

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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
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