Difference between revisions of "2023 AMC 10A Problems/Problem 14"
(→Solution 1) |
(→Solution 1) |
||
| Line 5: | Line 5: | ||
==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}. | + | 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}.\\$ |
| − | \\ | + | |
| − | 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\\ | ||
| Line 15: | Line 15: | ||
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 | + | 99 = 3^2 * 11: 11, 33, 99 - 1/2$ |
~vaisri | ~vaisri | ||
Revision as of 16:41, 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 $\boxed{B}.\$
11: 11 - 1/2$ (Error compiling LaTeX. Unknown error_msg)\\ 22 = 2 * 11: 11, 22 - 1/2\\ 33 = 3 * 11: 11, 33 - 1/2\\ 44 = 2^2 * 11: 11, 22, 44 - 1/2\\ 55 = 5 * 11: 11, 55 - 1/2\\ 66 = 2 * 3 * 11: 11, 22, 33, 66 - 1/2\\ 77 = 7 * 11: 11, 77 - 1/2\\ 88 = 2^3 * 11: 11, 22, 44, 88 - 1/2\\ 99 = 3^2 * 11: 11, 33, 99 - 1/2$
~vaisri
