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. | + | 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 has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
Solution
We note that the only thing that affects the parity of the factor are the powers of 2. There are factors of 2 in the number. Thus, there are 18 cases in which a factor of would be even (have a factor of in its prime factorization), and 1 case in which a factor of would be odd. Therefore, the answer is .
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
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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