2017 AMC 12B Problems/Problem 16

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 can consider a factor of $21!$ to be odd if it does not contain a $2$ ; hence, finding the exponent of $2$ in the prime factorization of $21!$ will help us find our answer. We can start off with all multiples of $2$ up to $21$ , which is $10$ . Then, we find multiples of $4$ , which is $5$ . Next, we look at multiples of $8$ , of which there are $2$ . Finally, we know there is only one multiple of $16$ in the set of positive integers up to $21$ . Now, we can add all of these to get $10+5+2+1=18$ . We know that, in the prime factorization of $21!$ , we have $2^{18}$ , and the only way to have an odd number is if there is not a $2$ in that number's prime factorization. This only happens with $2^{0}$ , which is only one of the 19 different exponents of 2 we could have (of which having each exponent is equally likely). Hence, we have $\boxed{\text{(B)} \dfrac{1}{19}}.$

Solution by: armang32324

Solution

If $21!$ prime factorizes into $p$ prime factors with exponents $e_1$ through $e_k$ , then the product of the sums of each of these exponents plus $1$ should be over $60,000$ . If $e_2$ is the exponent of $2$ in the prime factorization of 21!, then we can find the number of odd factors of $21!$ by dividing the total by $e_2+1$ . Then, the number of odd divisors over total divisors is $\dfrac{1}{e_2+1}$ . We can find $e_2$ easily using Legendre's, so our final answer is $\dfrac{1}{10 + 5 + 2 + 1+1} = \boxed{\text{(B)} \dfrac{1}{19}}.$

~ icecreamrolls8

Solution

If a factor of $21!$ is odd, that means it contains no factors of $2$ . We can find the number of factors of two in $21!$ by counting the number multiples of $2$ , $4$ , $8$ , and $16$ that are less than or equal to $21$ (Legendre's Formula). After some quick counting we find that this number is $10+5+2+1 = 18$ . If the prime factorization of $21!$ has $18$ factors of $2$ , there are $19$ choices for each divisor for how many factors of $2$ should be included ( $0$ to $18$ inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of $2$ is $0$ which is $\boxed{\textbf{(B)}\frac{1}{19}}$ .

Solution by: vedadehhc

Solution 2

We can write $21!$ as its prime factorization: $21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19$

Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; $2^{18}$ is going to have $19$ factors: $2^0, 2^1, 2^2,...\text{ }2^{18}$ , and the other exponents will behave identically.

In other words, $21!$ has $(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)$ factors.

We are looking for the probability that a randomly chosen factor of $21!$ will be odd--numbers that do not contain multiples of $2$ as factors.

From our earlier observation, the only factors of $21!$ that are even are ones with at least one multiplier of $2$ , so our probability of finding an odd factor becomes the following: $P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}$

Solution submitted by David Kim

Video Solution

https://youtu.be/ZLHNTSIcGM8

-MistyMathMusic

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

Page

Toolbox

Search

2017 AMC 12B Problems/Problem 16

Contents

Problem

Solution 1

Solution

Solution

Solution 2

Video Solution

See Also