Difference between revisions of "2023 AMC 10A Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
| − | As stated in Solution 1, the 9 multiples of 11 under <math>100</math> are 11, 22, 33, 44, 55, 66, 77, 88, 99. Because all of these numbers are multiples of 11 to the first power and first power only, their factors can either have 11 as a factor (<math>11^{1}</math>) or not have 11 as a factor (<math>11^{0}</math>), resulting in a 1/ | + | As stated in Solution 1, the <math>9</math> multiples of <math>11</math> under <math>100</math> are <math>11</math>, <math>22</math>, <math>33</math>, <math>44</math>, <math>55</math>, <math>66</math>, <math>77</math>, <math>88</math>, <math>99</math>. Because all of these numbers are multiples of <math>11</math> to the first power and first power only, their factors can either have <math>11</math> as a factor (<math>11^{1}</math>) or not have <math>11</math> as a factor (<math>11^{0}</math>), resulting in a <math>\frac{1}{2}</math> chance of a factor chosen being divisible by <math>11</math>. The chance of choosing any factor of <math>11</math> under <math>100</math> is <math>\frac{9}{100}</math>, so the final answer is <math>\frac{9}{100} \cdot \frac{1}{2} = \boxed{\textbf{(B)}~\frac{9}{200}}.</math> |
~Failure.net | ~Failure.net | ||
Revision as of 22:59, 11 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
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)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
