Difference between revisions of "2014 AMC 12B Problems/Problem 7"

Revision as of 12:23, 22 February 2014

Problem

For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}}\ 7\qquad\textbf{(E)}\ 8$ (Error compiling LaTeX. Unknown error_msg)

Solutions

Solution 1

We know that $n \le 30$ or else $30-n$ will be negative, resulting in a negative fraction. We also know that $n \ge 15$ or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value. Substituting all values $n$ from $15$ to $30$ gives us integer values for $n=15, 20, 24, 25, 27, 28, 29$ . Counting them up, we have $\boxed{\textbf{(D)}\ 7}$ possible values for $n$ .

Solution 2

Let $\frac{n}{30-n}=m$ , where $m \in \mathbb{N}$ . Solving for $n$ , we find that $n=\frac{30m}{m+1}$ . Because $m$ and $m+1$ are relatively prime, $m+1|30$ . Our answer is the number of proper divisors of $2^13^15^1$ , which is $(1+1)(1+1)(1+1)-1 = \boxed{\textbf{(D)}\ 7}$ .

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 13: / Line 13: @@
 Let <math> \frac{n}{30-n}=m </math>, where <math> m \in \mathbb{N} </math>. Solving for <math> n </math>, we find that <math> n=\frac{30m}{m+1} </math>. Because <math> m </math> and <math> m+1 </math> are relatively prime, <math> m+1|30 </math>. Our answer is the number of proper divisors of <math> 2^13^15^1 </math>, which is <math> (1+1)(1+1)(1+1)-1 = \boxed{\textbf{(D)}\ 7} </math>.
+== See also ==
 {{AMC12 box|year=2014|ab=B|num-b=6|num-a=8}}
 {{MAA Notice}}

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