Difference between revisions of "2023 AMC 10A Problems/Problem 14"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/N2lyYRMuZuk?si=-CyrdswJABoMrnWk&t=825 | ||
+ | ~Math-X | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2023|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 08:37, 12 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 ?
Contents
[hide]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
Solution 2
As stated in Solution 1, the multiples of under are , , , , , , , , . 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
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Math-X (First understand the problem!!!)
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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