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

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==Solution==
 
==Solution==
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>\frac{1}{19} \Rightarrow \boxed{(B)}</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>.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 20:46, 16 February 2017

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 $21$.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

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