Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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Revision as of 21:56, 16 February 2017
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
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. 
