2014 AMC 12B Problems/Problem 7
Contents
[hide]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
.
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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