Difference between revisions of "2017 AMC 12B Problems/Problem 16"

Latest revision as of 23:43, 8 November 2024

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 2

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 3

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

Note: Legendre's Formula states that the exponent of $p$ in the prime factorization of $n!$ is given by $e_p(n!)=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}\right\rfloor=\frac{n-S_p(n)}{p-1},$ where $S_p(n)$ is the sum of the digits of $n$ when written in base $p$ . For example, $21$ in binary is $1011$ , so $S_2(21)=1+0+1+1=3$ . We have $e_2(21!)=\frac{21-3}{2-1}=18$ , then proceed with the rest of solution 3.

~ happyhippos

Solution 4

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

@@ Line 16: / Line 16: @@
 ~ icecreamrolls8
-==Solution==
+==Solution 3==
 If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math> (Legendre's Formula). After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is     <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>.
 Solution by: vedadehhc
-==Solution 2==
+Note: Legendre's Formula states that the exponent of <math>p</math> in the prime factorization of <math>n!</math> is given by
+<cmath>e_p(n!)=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}\right\rfloor=\frac{n-S_p(n)}{p-1},</cmath>
+where <math>S_p(n)</math> is the sum of the digits of <math>n</math> when written in base <math>p</math>. For example, <math>21</math> in binary is <math>1011</math>, so <math>S_2(21)=1+0+1+1=3</math>. We have <math>e_2(21!)=\frac{21-3}{2-1}=18</math>, then proceed with the rest of solution 3.
+~ happyhippos
+==Solution 4==
 We can write <math>21!</math> as its prime factorization:
 <cmath>21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19</cmath>

2017 AMC 12B (Problems • Answer Key • Resources)
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