Difference between revisions of "2017 AMC 10B Problems/Problem 20"
m (→Solution 2 (Constructive Counting)) |
m (→Solution 1) |
||
| Line 4: | Line 4: | ||
<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 1== | ==Solution 1== | ||
| − | We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are <math>18</math> cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and <math>1</math> case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math> | + | We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are <math>18</math> cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and <math>1</math> case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math>. |
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. | 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. | ||
Revision as of 18:11, 17 January 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= $\boxed{\text{(B)} \frac{1,19}}$ (Error compiling LaTeX. Unknown error_msg)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
