2017 AMC 12B Problems/Problem 16
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[hide]Problem
The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
Solution 1
We can consider a factor of to be odd if it does not contain a ; hence, finding the exponent of in the prime factorization of will help us find our answer. We can start off with all multiples of up to , which is . Then, we find multiples of , which is . Next, we look at multiples of , of which there are . Finally, we know there is only one multiple of in the set of positive integers up to . Now, we can add all of these to get . We know that, in the prime factorization of , we have , and the only way to have an odd number is if there is not a in that number's prime factorization. This only happens with , which is only one of the 19 different exponents of 2 we could have (of which having each exponent is equally likely). Hence, we have
Solution by: armang32324
Solution 2
If prime factorizes into prime factors with exponents through , then the product of the sums of each of these exponents plus should be over . If is the exponent of in the prime factorization of 21!, then we can find the number of odd factors of by dividing the total by . Then, the number of odd divisors over total divisors is . We can find easily using Legendre's, so our final answer is
~ icecreamrolls8
Solution 3
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
Note: Legendre's Formula states that the exponent of in the prime factorization of is given by where is the sum of the digits of when written in base . For example, in binary is , so . We have , then proceed with the rest of solution 3.
~ happyhippos
Solution 4
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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