2023 AMC 10A Problems/Problem 14
Contents
Problem
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
In order for the divisor chosen to be a multiple of , the original number chosen must also be a multiple of . 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 ~(minor formatting changes by Marshall_Huang)
Solution 2
There are multiples of under . Because all of these numbers are multiples of to the first power and first power only, their factors can either have as a factor () or not have as a factor (), resulting in a chance of a factor chosen being divisible by . The chance of choosing any factor of under is , so the final answer is
~Failure.net
Solution 3 (generalized)
Let be the positive integer in question. Since is a multiple of , we can write the prime factorization of . To find the total number of divisors divisible by , observe that there are divisors not divisible by . For each power of greater than , of which there are , it can be paired with any of these other divisors. Therefore, there are divisors of that are divisible by . The total number of divisors of is so the probability of choosing a divisor that is divisible by is For each positive integer less than or equal to , the highest power of that divides it is , so , meaning that the probability of choosing a divisor (that is divisible by ) of a fixed is The probability of choosing any from the first positive integers is so the probability of choosing any of these divisors is There are multiples of less than or equal to , so the total probability is
-Benedict T (countmath1)
Video Solutions
https://www.youtube.com/watch?v=jkfsBYzBJbQ
https://www.youtube.com/watch?v=uaf46N6qP54
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
https://youtu.be/N2lyYRMuZuk?si=-CyrdswJABoMrnWk&t=825
~Math-X
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 |
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