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

Revision as of 18:12, 20 August 2019

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

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

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.

@@ Line 25: / Line 25: @@
 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 0, 1, 2, 3 as the possibilities. <math>\boxed{(D)}</math>
+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

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