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

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==Solution==
 
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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>.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>.
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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 by: vedadehhc

Revision as of 17:22, 4 February 2020

Problem 16

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

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

See Also

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

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