2017 AMC 12B Problems/Problem 16
Problem 16
The number has over
positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
Solution
If a factor of is odd, that means it contains no factors of
. We can find the number of factors of two in
by counting the number multiples of
,
,
, and
that are less than or equal to
.After some quick counting we find that this number is
. If the prime factorization of
has
factors of
, there are
choices for each divisor for how many factors of
should be included (
to
inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of
is
which is
.
Solution by: vedadehhc
Solution 2
We can write as its prime factorization:
Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; is going to have
factors:
, and the other exponents will behave identically.
In other words, has
factors.
We are looking for the probability that a randomly chosen factor of will be odd--numbers that do not contain multiples of
as factors.
From our earlier observation, the only factors of that are even are ones with at least one multiplier of
, so our probability of finding an odd factor becomes the following:
Solution submitted by David Kim
Solution 3 (Time Shortage)
: Only use this solution if you are running out of time.
Prime factor . It is
.
The number of factors is .
Looking at the choices and using some quick computation, one can figure that the only sensical answer is
.
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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