2017 AMC 12B Problems/Problem 16
Problem
The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
Solution
If prime factorizes into prime factors with exponents through , then the product of the sums of each of these exponents plus should be . If we divide this product by the exponent of plus , then we should get the number of odd factors. Then, the fraction of odd divisors over total divisors is if is the exponent of . We can find easily using Legendre's, so our final answer is
~ icecreamrolls8
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
-MistyMathMusic
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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