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>. | ||
+ | |||
+ | ===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> | ||
== 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 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 .
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 |
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