Difference between revisions of "2014 AMC 12B Problems/Problem 7"
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For how many positive integers <math>n</math> is <math>\frac{n}{30-n}</math> also a positive integer? | For how many positive integers <math>n</math> is <math>\frac{n}{30-n}</math> also a positive integer? | ||
− | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D) | + | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math> |
==Solutions== | ==Solutions== | ||
Line 10: | Line 10: | ||
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>. | 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 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>. | ||
− | ===Solution | + | ==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 == | ||
+ | {{AMC12 box|year=2014|ab=B|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:52, 14 August 2024
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
Solution 4
We recognize that because positive integer, it is easy to just test the numbers, yielding:
29, 28, 27, 25, 24, 20, 15
meaning we have
~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 |
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