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

Revision as of 17:46, 20 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 by kevin38017)

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, $m|30$ . Our answer is the number of proper divisors of $2^13^15^1$ , which is $(1+1)(1+1)(1+1)-1 = \boxed{\textbf{(D)}\ 7}$ .

@@ Line 5: / Line 5: @@
 <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}}\ 7\qquad\textbf{(E)}\ 8 </math>
-==Solution==
+==Solutions==
+===Solution 1===
 We know that <math>n \le 30</math> or else <math>30-n</math> will be negative, resulting in a negative fraction.  We also know that <math>n \ge 15</math> or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value.  Substituting all values <math>n</math> from <math>15</math> to <math>30</math> gives us integer values for <math>n=15, 20, 24, 25, 27, 28, 29</math>.  Counting them up, we have <math>\boxed{\textbf{(D)}\ 7}</math> possible values for <math>n</math>.
 (Solution by kevin38017)
+===Solution 2===
+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|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>.

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