Difference between revisions of "2017 AMC 10B Problems/Problem 20"
(→Solution 2: Constructive counting) |
|||
Line 7: | Line 7: | ||
==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), 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> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:00, 12 August 2017
Problem
The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
Solution
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=
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.