Difference between revisions of "2017 AMC 10B Problems/Problem 20"
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Consider how to construct any divisor <math>D</math> of <math>21!</math>. First by Legendre's theorem for the divisors of a factorial (see here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. <math>D</math> can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for <math>D</math> to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases = <math>\boxed{\textbf{(B)} \frac 1{19}}</math> | Consider how to construct any divisor <math>D</math> of <math>21!</math>. First by Legendre's theorem for the divisors of a factorial (see here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. <math>D</math> can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for <math>D</math> to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases = <math>\boxed{\textbf{(B)} \frac 1{19}}</math> | ||
==Solution 3== | ==Solution 3== | ||
| − | We can find the prime factorization of 21!. We do this by finding the prime factorization of each of 21, 20, ..., 2, and 1 and multiplying them together. This gives us <math>2^{18}</math> x <math>3^{9}</math> x <math>5^{4}</math> x <math>7^{3}</math> x 11 x 13 x 17 x 19. To find the number of odd divisors, we pretend as if the <math>2^{18}</math> doesn't exist and count the other divisors: 10 x 5 x 4 x 2 x 2 x 2 x 2. The total number of divisors are 19 x 10 x 5 x 4 x 2 x 2 x 2 x 2. Dividing gives ( | + | We can find the prime factorization of 21!. We do this by finding the prime factorization of each of 21, 20, ..., 2, and 1 and multiplying them together. This gives us <math>2^{18}</math> x <math>3^{9}</math> x <math>5^{4}</math> x <math>7^{3}</math> x 11 x 13 x 17 x 19. To find the number of odd divisors, we pretend as if the <math>2^{18}</math> doesn't exist and count the other divisors: 10 x 5 x 4 x 2 x 2 x 2 x 2. The total number of divisors are 19 x 10 x 5 x 4 x 2 x 2 x 2 x 2. Dividing gives (B) 1/19. |
==See Also== | ==See Also== | ||
Revision as of 15:38, 7 November 2021
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 note that the only thing that affects the parity of the factor are the powers of 2. There are 
factors of 2 in the number. Thus, there are 
cases in which a factor of 
would be even (have a factor of 
in its prime factorization), and 
case in which a factor of would be odd. Therefore, the answer is 
.
Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21!into 2 groups, one with powers of 2 and the others odd factors. For each power of 2, the factors combine a certain number of 2's from the first group and numbers from the odd group. That is why symmetry occurs here.
Solution 2 (Constructive Counting)
Consider how to construct any divisor 
of . First by Legendre's theorem for the divisors of a factorial (see here: Legendre's Formula), we have that there are a total of 18 factors of 2 in the number. can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases =
Solution 3
We can find the prime factorization of 21!. We do this by finding the prime factorization of each of 21, 20, ..., 2, and 1 and multiplying them together. This gives us 
x 
x 
x 
x 11 x 13 x 17 x 19. To find the number of odd divisors, we pretend as if the doesn't exist and count the other divisors: 10 x 5 x 4 x 2 x 2 x 2 x 2. The total number of divisors are 19 x 10 x 5 x 4 x 2 x 2 x 2 x 2. Dividing gives (B) 1/19.
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 10 Problems and Solutions | ||
| 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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