2017 AMC 10B Problems/Problem 20

Problem

The number $21!=51,090,942,171,709,440,000$ has over $60,000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}$

Solution 1

We note that the only thing that affects the parity of the factor are the powers of 2. There are $10+5+2+1 = 18$ factors of 2 in the number. Thus, there are $18$ cases in which a factor of $21!$ would be even (have a factor of $2$ in its prime factorization), and $1$ case in which a factor of $21!$ would be odd. Therefore, the answer is $\boxed{\textbf{(B)} \frac 1{19}}$

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 $D$ of $21!$ . 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. $D$ can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for $D$ to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases= $\boxed{1/19, \space \text{B}}$

Solution 3: Solve out number of divisors, then use complimentary counting

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...)

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 10 Problems and Solutions

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 12 Problems and Solutions

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2017 AMC 10B Problems/Problem 20

Contents

Problem

Solution 1

Solution 2: Constructive counting

Solution 3: Solve out number of divisors, then use complimentary counting

See Also