Difference between revisions of "2023 AMC 10A Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | Among the first <math>100</math> positive integers, there are 9 multiples of 11. 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 <math>\frac{9}{100}</math>, so the final probability is <math>\frac{9}{100} \cdot \frac{1}{2} = \frac{9}{200}</math>, so the answer is <math>\boxed{B}.</math> | + | Among the first <math>100</math> 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 <math>\frac{9}{100}</math>, so the final probability is <math>\frac{9}{100} \cdot \frac{1}{2} = \frac{9}{200}</math>, so the answer is <math>\boxed{B}.</math> |
<math>11 = 11 - 1/2\ | <math>11 = 11 - 1/2\ | ||
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99 = 3^2 * 11: 11, 33, 99 - 1/2</math> | 99 = 3^2 * 11: 11, 33, 99 - 1/2</math> | ||
− | ~vaisri | + | ~vaisri ~walmartbrian ~Shontai |
Revision as of 16:57, 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