Difference between revisions of "2017 AMC 10B Problems/Problem 20"
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==Problem== | ==Problem== | ||
| − | + | The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd? | |
| + | <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>. | 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>. | ||
Revision as of 13:16, 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. Since there are 
factors of 2, and since only haveing a 0 power of 2 makes the factor odd, then 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
