Difference between revisions of "2023 AMC 10A Problems/Problem 14"
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Revision as of 20:25, 9 November 2023
A number is chosen at random from among the first 
positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by 
?
Solution 1
Among the first positive integers, there are 9 multiples of 11; 11, 22, 33, 44, 55, 66, 77, 88, 99. We can now perform a little casework on the probability of choosing a divisor which is a multiple of 11 for each of these 9, and see that the probability is 1/2 for each. The probability of choosing these 9 multiples in the first place is 
, so the final probability is 
, so the answer is 
~vaisri ~walmartbrian ~Shontai
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
