# 2014 AMC 12B Problems/Problem 7

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## 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)

~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 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

## See also

 2014 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 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

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