Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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Solution submitted by [[User:TrueshotBarrage|David Kim]] | Solution submitted by [[User:TrueshotBarrage|David Kim]] | ||
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+ | ==Video Solution== | ||
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+ | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_16 | ||
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+ | -MistyMathMusic | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}} |
Revision as of 22:17, 11 August 2020
Contents
[hide]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
(Legendre's Formula). 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
Video Solution
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_16
-MistyMathMusic
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.