2017 AMC 10B Problems/Problem 20
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: http://www.cut-the-knot.oIlicktoesrg/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 |
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.