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>. | ||
+ | |||
+ | ==Video Solution 1 (Quick and Easy)== | ||
+ | https://youtu.be/rN76FKYRjls | ||
+ | |||
+ | ~Education, the Study of Everything | ||
===Solution 3=== | ===Solution 3=== |
Revision as of 00:38, 15 November 2022
Contents
Problem
For how many positive integers is
also a positive integer?
Solutions
Solution 1
We know that or else
will be negative, resulting in a negative fraction. We also know that
or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value. Substituting all values
from
to
gives us integer values for
. Counting them up, we have
possible values for
.
Solution 2
Let , where
. Solving for
, we find that
. Because
and
are relatively prime,
. Our answer is the number of proper divisors of
, which is
.
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Solution 3
We know that . Then, by divisibility rules:
There are divisors of
, but
must be positive, so
isn't counted, meaning we have
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.