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

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===Solution 2===
 
===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+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>.
 
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>.
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===Solution 3===
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We know that <math>30-n|n</math>. Then, by divisibility rules:
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<cmath>\Leftrightarrow 30-n|n+30-n</cmath>
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<cmath>\Leftrightarrow 30-n|30</cmath>
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There are <math>8</math> divisors of <math>30</math>, but <math>n</math> must be positive, so <math>30|30</math> isn't counted, meaning we have <math>\boxed{\textbf{(D)}\ 7} </math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2014|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2014|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:05, 2 February 2021

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

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 $\boxed{\textbf{(D)}\ 7}$

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
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

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