Difference between revisions of "2017 AMC 10B Problems/Problem 20"
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==Solution 2: Constructive counting== | ==Solution 2: Constructive counting== | ||
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: http://www.cut-the-knot.org/blue/LegendresTheorem.shtml and 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{1/19, \space \text{B}}</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: http://www.cut-the-knot.org/blue/LegendresTheorem.shtml and 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{1/19, \space \text{B}}</math> | ||
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| + | ==Solution 3: | ||
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| + | Since there are 18 powers of two, we see that out of the total cases, 18/19 of them will be even, so only 1/19 of them will be odd (similar to solution 1, but solved the whole thing out...) | ||
==See Also== | ==See Also== | ||
Revision as of 18:31, 2 January 2020
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 
Solution 2: Constructive counting
Consider how to construct any divisor 
of . First by Legendre's theorem for the divisors of a factorial (see here: http://www.cut-the-knot.org/blue/LegendresTheorem.shtml and 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:
Since there are 18 powers of two, we see that out of the total cases, 18/19 of them will be even, so only 1/19 of them will be odd (similar to solution 1, but solved the whole thing out...)
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. 
