Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math> (Legendre's Formula). After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>. | If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math> (Legendre's Formula). After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>. | ||
Revision as of 22:03, 30 September 2024
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
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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