Difference between revisions of "2017 AMC 10B Problems/Problem 20"

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<math>\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}</math>
 
<math>\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}</math>
 
==Solution==
 
==Solution==
We note that the only thing that affects the parity of the factor are the powers of 2. Since there are <math>10+5+2+1 = 18</math> factors of 2, and since only haveing a 0 power of 2 makes the factor odd, then the answer is <math>\boxed{\textbf{(B) } \frac 1{19}}</math>.
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We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are 18 cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and 1 case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:24, 16 February 2017

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

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 $\boxed{\textbf{(B)} \frac 1{19}}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 10 Problems and Solutions

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