Difference between revisions of "2023 AMC 10A Problems/Problem 14"
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− | 11: 11 - 1/2<math>\ | + | 11: 11 - 1/2<math>\\ |
− | 22 = 2 * 11: 11, 22 - 1/2 | + | 22 = 2 * 11: 11, 22 - 1/2\\ |
− | 33 = 3 * 11: 11, 33 - 1/2 | + | 33 = 3 * 11: 11, 33 - 1/2\\ |
− | 44 = 2^2 * 11: 11, 22, 44 - 1/2 | + | 44 = 2^2 * 11: 11, 22, 44 - 1/2\\ |
− | 55 = 5 * 11: 11, 55 - 1/2 | + | 55 = 5 * 11: 11, 55 - 1/2\\ |
− | 66 = 2 * 3 * 11: 11, 22, 33, 66 - 1/2 | + | 66 = 2 * 3 * 11: 11, 22, 33, 66 - 1/2\\ |
− | 77 = 7 * 11: 11, 77 - 1/2 | + | 77 = 7 * 11: 11, 77 - 1/2\\ |
− | 88 = 2^3 * 11: 11, 22, 44, 88 - 1/2 | + | 88 = 2^3 * 11: 11, 22, 44, 88 - 1/2\\ |
− | 99 = 3^2 * 11: 11, 33, 99 - 1/2<math>\ | + | 99 = 3^2 * 11: 11, 33, 99 - 1/2</math>\\ |
~vaisri | ~vaisri |
Revision as of 15:40, 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. 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
Casework: 11: 11 - 1/2\
~vaisri