Difference between revisions of "2023 AMC 10A Problems/Problem 14"
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<math>\textbf{(A)}~\frac{4}{100}\qquad\textbf{(B)}~\frac{9}{200} \qquad \textbf{(C)}~\frac{1}{20} \qquad\textbf{(D)}~\frac{11}{200}\qquad\textbf{(E)}~\frac{3}{50}</math> | <math>\textbf{(A)}~\frac{4}{100}\qquad\textbf{(B)}~\frac{9}{200} \qquad \textbf{(C)}~\frac{1}{20} \qquad\textbf{(D)}~\frac{11}{200}\qquad\textbf{(E)}~\frac{3}{50}</math> | ||
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| + | ==Solution 1== | ||
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| + | 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}.</math> | ||
Revision as of 15:34, 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 $\frac{9}{100} \cdot \frac{1{2} = \frac{9}{200}$ (Error compiling LaTeX. Unknown error_msg), so the answer is 
