2023 AMC 10A Problems/Problem 14

Problem

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

In order for the divisor chosen to be a multiple of $11$ , the original number chosen must also be a multiple of $11$ . Among the first $100$ 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 $\frac{9}{100}$ , so the final probability is $\frac{9}{100} \cdot \frac{1}{2} = \frac{9}{200}$ , so the answer is $\boxed{\textbf{(B)}~\frac{9}{200}}.$

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

~vaisri ~walmartbrian ~Shontai

Solution 2

As stated in Solution 1, the $9$ multiples of $11$ under $100$ 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 ( $11^{1}$ ) or not have $11$ as a factor ( $11^{0}$ ), resulting in a $\frac{1}{2}$ chance of a factor chosen being divisible by $11$ . The chance of choosing any factor of $11$ under $100$ is $\frac{9}{100}$ , so the final answer is $\frac{9}{100} \cdot \frac{1}{2} = \boxed{\textbf{(B)}~\frac{9}{200}}.$

~Failure.net

Video Solution

https://youtu.be/O_P3E9hDrYY

~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

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
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