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Difference between revisions of "2023 AMC 10A Problems/Problem 14"

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(→‎Solution 1)
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Casework:
 
Casework:
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11: 11 - 1/2<math>\newline</math>
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11: 11 - 1/2<math>\\
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22 = 2 * 11: 11, 22 - 1/2<math>\newline</math>
+
22 = 2 * 11: 11, 22 - 1/2\\
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33 = 3 * 11: 11, 33 - 1/2<math>\newline</math>
+
33 = 3 * 11: 11, 33 - 1/2\\
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44 = 2^2 * 11: 11, 22, 44 - 1/2<math>\newline</math>
+
44 = 2^2 * 11: 11, 22, 44 - 1/2\\
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55 = 5 * 11: 11, 55 - 1/2<math>\newline</math>
+
55 = 5 * 11: 11, 55 - 1/2\\
−
66 = 2 * 3 * 11: 11, 22, 33, 66 - 1/2<math>\newline</math>
+
66 = 2 * 3 * 11: 11, 22, 33, 66 - 1/2\\
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77 = 7 * 11: 11, 77 - 1/2<math>\newline</math>
+
77 = 7 * 11: 11, 77 - 1/2\\
−
88 = 2^3 * 11: 11, 22, 44, 88 - 1/2<math>\newline</math>
+
88 = 2^3 * 11: 11, 22, 44, 88 - 1/2\\
−
99 = 3^2 * 11: 11, 33, 99 - 1/2<math>\newline</math>
+
99 = 3^2 * 11: 11, 33, 99 - 1/2</math>\\
  
 
~vaisri
 
~vaisri

Revision as of 15:40, 9 November 2023

A number is chosen at random from among the first $100$ 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 $11$?

$\textbf{(A)}~\frac{4}{100}\qquad\textbf{(B)}~\frac{9}{200} \qquad \textbf{(C)}~\frac{1}{20} \qquad\textbf{(D)}~\frac{11}{200}\qquad\textbf{(E)}~\frac{3}{50}$

Solution 1

Among the first $100$ 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 $\frac{9}{100}$, so the final probability is $\frac{9}{100} \cdot \frac{1}{2} = \frac{9}{200}$, so the answer is $\boxed{B}.$

Casework: 11: 11 - 1/2$\\ 22 = 2 * 11: 11, 22 - 1/2\\ 33 = 3 * 11: 11, 33 - 1/2\\ 44 = 2^2 * 11: 11, 22, 44 - 1/2\\ 55 = 5 * 11: 11, 55 - 1/2\\ 66 = 2 * 3 * 11: 11, 22, 33, 66 - 1/2\\ 77 = 7 * 11: 11, 77 - 1/2\\ 88 = 2^3 * 11: 11, 22, 44, 88 - 1/2\\ 99 = 3^2 * 11: 11, 33, 99 - 1/2$\\

~vaisri

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