2014 AMC 12B Problems/Problem 7
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 |
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All AMC 12 Problems and Solutions |
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