Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Solution 3)
 
(2 intermediate revisions by 2 users not shown)
Line 12: Line 12:
 
===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>.
 +
 +
==Video Solution 1 (Quick and Easy)==
 +
https://youtu.be/rN76FKYRjls
 +
 +
~Education, the Study of Everything
 +
 +
===Solution 3===
 +
We know that <math>30-n|n</math>. Then, by divisibility rules:
 +
 +
<cmath>\Leftrightarrow 30-n|n+30-n</cmath>
 +
<cmath>\Leftrightarrow 30-n|30</cmath>
 +
 +
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>
 +
===Solution 4===
 +
We recognize that <math>15<n<30</math> because positive integer, it is easy to just test the numbers, yielding:
 +
 +
29, 28, 27, 25, 24, 20, 15
 +
 +
meaning we have <math>\boxed{\textbf{(D)}\ 7} </math>
 +
~MathCosine
  
 
== 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 15:52, 14 August 2024

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

See also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_7&oldid=225023"