Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12B Problems/Problem 12"
(→‎Solution 2)
 
(17 intermediate revisions by 8 users not shown)
Line 8: Line 8:
 
\qquad\mathrm{(D)}\ 4
 
\qquad\mathrm{(D)}\ 4
 
\qquad\mathrm{(E)}\ 10</math>
 
\qquad\mathrm{(E)}\ 10</math>
−
== Solution ==
+
== Solution 1 ==
  
−
=== Solution 1 ===
+
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>\boxed{\mathrm{(D)}\ 4}</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>).
+
== Solution 2 ==
−
 
 
−
=== Solution 2 ===
 
  
 
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.
 
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.
Line 24: Line 22:
 
For prime <math>n</math> the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
 
For prime <math>n</math> the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
  
−
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.
+
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. Therefore, there are only <math>\boxed{\mathrm{(D)}\ 4}</math> solutions (respectively yielding <math>n = 0, 10, 16, 18</math>).
 +
 
 +
== Solution 3 ==
 +
 
 +
If <math>\frac{n}{20-n} = k^2 \ge 0</math>, then <math>n \ge 0</math> and <math>20-n > 0</math>, otherwise <math>\frac{n}{20-n}</math> will be negative. Thus <math>0 \le n \le 19</math> and <cmath>0 = \frac{0}{20-(0)} \le \frac{n}{20-n} \le \frac{19}{20-(19)} = 19</cmath> Checking all <math>k</math> for which <math>0 \le k^2 \le 19</math>, we have <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math> as the possibilities. <math>\boxed{(D)}</math>
 +
 
 +
~ Nafer
 +
 
 +
== Solution 4 ==
 +
For all integers x, <math>x^2</math> is always a positive integer. So solve for <math>\frac{n}{20-n} = 0</math>, getting <math>n=0</math> and <math>\frac{n}{20-n} = 1</math>, getting <math>n
 +
=10</math>. For all values n less than 0 and greater than 20, the value <math>\frac{n}{20-n}</math> is negative, so now try values of n between 10 and 20. Quick substitution finds <math>0</math>, <math>10</math>, <math>16</math>, and <math>18</math> which yields <math>x=0</math>, <math>x=1</math>, <math>x=2</math>, and <math>x=3</math> respectively. 4 values, or <math>\boxed{(D)}</math>
 +
 
 +
 
 +
== Solution 5 ==
 +
Simon's Favourite Factoring Trick.
 +
 
 +
Since <math>\frac{n}{20-n}</math> is an integer <math>k</math>, we multiply both sides by <math>20-n</math>. This gives us <math>n=20k^2</math>-<math>nk^2</math>. We subtract <math>20k^2</math> on both sides, then add <math>nk^2</math> on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have <math>(k^2+1)(n-20)=20</math>. We then consider the different factors of <math>20</math> that <math>k^2+1</math> can be. It could be <math>1,2,4,5,10</math>, and <math>20</math>. After checking case by case, we then are able to identify that there are <math>4</math> such <math>k</math> values that also yield an integer <math>n</math> value, meaning that there are <math>4</math> values, so the correct answer is  <math>\boxed{(D)}</math>
 +
~CharmaineMa07292010
  
 
== See also ==
 
== See also ==

Latest revision as of 19:26, 30 October 2024

The following problem is from both the 2002 AMC 12B #12 and 2002 AMC 10B #16, so both problems redirect to this page.

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 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 $\boxed{\mathrm{(D)}\ 4}$ 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. Therefore, there are only $\boxed{\mathrm{(D)}\ 4}$ solutions (respectively yielding $n = 0, 10, 16, 18$).

Solution 3

If $\frac{n}{20-n} = k^2 \ge 0$, then $n \ge 0$ and $20-n > 0$, otherwise $\frac{n}{20-n}$ will be negative. Thus $0 \le n \le 19$ and \[0 = \frac{0}{20-(0)} \le \frac{n}{20-n} \le \frac{19}{20-(19)} = 19\] Checking all $k$ for which $0 \le k^2 \le 19$, we have $0$, $1$, $2$, $3$ as the possibilities. $\boxed{(D)}$

~ Nafer

Solution 4

For all integers x, $x^2$ is always a positive integer. So solve for $\frac{n}{20-n} = 0$, getting $n=0$ and $\frac{n}{20-n} = 1$, getting $n =10$. For all values n less than 0 and greater than 20, the value $\frac{n}{20-n}$ is negative, so now try values of n between 10 and 20. Quick substitution finds $0$, $10$, $16$, and $18$ which yields $x=0$, $x=1$, $x=2$, and $x=3$ respectively. 4 values, or $\boxed{(D)}$


Solution 5

Simon's Favourite Factoring Trick.

Since $\frac{n}{20-n}$ is an integer $k$, we multiply both sides by $20-n$. This gives us $n=20k^2$-$nk^2$. We subtract $20k^2$ on both sides, then add $nk^2$ on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have $(k^2+1)(n-20)=20$. We then consider the different factors of $20$ that $k^2+1$ can be. It could be $1,2,4,5,10$, and $20$. After checking case by case, we then are able to identify that there are $4$ such $k$ values that also yield an integer $n$ value, meaning that there are $4$ values, so the correct answer is $\boxed{(D)}$ ~CharmaineMa07292010

See also

2002 AMC 10B (Problems • Answer Key • Resources)
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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_12&oldid=231038"