Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math> | <math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math> | ||
− | ==Solution== | + | ==Solution 1== |
We can consider a factor of <math>21!</math> to be odd if it does not contain a <math>2</math>; hence, finding the exponent of <math>2</math> in the prime factorization of <math>21!</math> will help us find our answer. We can start off with all multiples of <math>2</math> up to <math>21</math>, which is <math>10</math>. Then, we find multiples of <math>4</math>, which is <math>5</math>. Next, we look at multiples of <math>8</math>, of which there are <math>2</math>. Finally, we know there is only one multiple of <math>16</math> in the set of positive integers up to <math>21</math>. Now, we can add all of these to get <math>10+5+2+1=18</math>. We know that, in the prime factorization of <math>21!</math>, we have <math>2^{18}</math>, and the only way to have an odd number is if there is not a <math>2</math> in that number's prime factorization. This only happens with <math>2^{0}</math>, 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 <math>\dfrac{1}{1+18} = \boxed{\text{(B)} \dfrac{1}{19}}.</math> | We can consider a factor of <math>21!</math> to be odd if it does not contain a <math>2</math>; hence, finding the exponent of <math>2</math> in the prime factorization of <math>21!</math> will help us find our answer. We can start off with all multiples of <math>2</math> up to <math>21</math>, which is <math>10</math>. Then, we find multiples of <math>4</math>, which is <math>5</math>. Next, we look at multiples of <math>8</math>, of which there are <math>2</math>. Finally, we know there is only one multiple of <math>16</math> in the set of positive integers up to <math>21</math>. Now, we can add all of these to get <math>10+5+2+1=18</math>. We know that, in the prime factorization of <math>21!</math>, we have <math>2^{18}</math>, and the only way to have an odd number is if there is not a <math>2</math> in that number's prime factorization. This only happens with <math>2^{0}</math>, 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 <math>\dfrac{1}{1+18} = \boxed{\text{(B)} \dfrac{1}{19}}.</math> | ||
Solution by: armang32324 | Solution by: armang32324 | ||
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==Solution== | ==Solution== | ||
Revision as of 15:14, 6 April 2023
Contents
[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
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 we divide this product by the exponent of in 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 |
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 |
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