2014 AMC 12B Problems/Problem 7

Problem

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

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

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 $(1+1)(1+1)(1+1)-1 = \boxed{\textbf{(D)}\ 7}$ .

Video Solution 1 (Quick and Easy)

https://youtu.be/rN76FKYRjls

~Education, the Study of Everything

Solution 3

We know that $30-n|n$ . Then, by divisibility rules:

$\Leftrightarrow 30-n|n+30-n$ $\Leftrightarrow 30-n|30$

There are $8$ divisors of $30$ , but $n$ must be positive, so $30|30$ isn't counted, meaning we have $\boxed{\textbf{(D)}\ 7}$

Solution 4

We recognize that $15<n<30$ because positive integer, it is easy to just test the numbers, yielding:

29, 28, 27, 25, 24, 20, 15

meaning we have $\boxed{\textbf{(D)}\ 7}$ ~MathCosine

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

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

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