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

Revision as of 22:02, 10 May 2024

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

Solution 3 (generalized)

Let $N$ be the positive integer in question. Since $N$ is a multiple of $11$ , we can write $N = 11^{e_1}p_1^{e_2}\cdots p_{k-1}^{e_k},$ the prime factorization of $N$ . To find the total number of divisors divisible by $11$ , observe that there are $(e_2 + 1)(e_3 + 1)\cdots (e_k + 1)$ divisors not divisible by $11$ . For each power of $11$ greater than $1$ , of which there are $e_1$ , it can be paired with any of these other divisors. Therefore, there are $e_1(e_2+1)(e_3 + 1)\cdots (e_k + 1)$ divisors of $N$ that are divisible by $11$ . The total number of divisors of $11$ is $(e_1 + 1)(e_2 + 1)\cdots (e_k + 1),$ so the probability of choosing a divisor that is divisible by $11$ is $\frac{e_1(e_2+1)(e_3 + 1)\cdots (e_k + 1)}{(e_1 + 1)(e_2 + 1)\cdots (e_k + 1)} = \frac{e_1}{e_1 + 1}.$ For each positive integer less than or equal to $100$ , the highest power of $11$ that divides it is $11$ , so $e_1 = 1$ , meaning that the probability of choosing a divisor (that is divisible by $11$ ) of a fixed $N$ is $\frac{1}{2}.$ The probability of choosing any $N$ from the first $100$ positive integers is $\frac{1}{100},$ so the probability of choosing any of these divisors is $\frac{1}{100}\cdot \frac{1}{2}.$ There are $9$ multiples of $11$ less than or equal to $100$ , so the total probability is $9\cdot \frac{1}{100}\cdot \frac{1}{2} = \boxed{\textbf{(B)}\ \frac{9}{200}}.$

-Benedict T (countmath1)

Video Solution by Power Solve (easy to digest!)

https://www.youtube.com/watch?v=jkfsBYzBJbQ

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=uaf46N6qP54

Video Solution

https://youtu.be/O_P3E9hDrYY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

https://youtu.be/N2lyYRMuZuk?si=-CyrdswJABoMrnWk&t=825 ~Math-X

@@ Line 50: / Line 50: @@
 ~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
+~Math-X
 ==See Also==
 {{AMC10 box|year=2023|ab=A|num-b=13|num-a=15}}
 {{MAA Notice}}

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