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

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==Problem==
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A number is chosen at random from among the first <math>100</math> 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 <math>11</math>?
 
A number is chosen at random from among the first <math>100</math> 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 <math>11</math>?
  
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==Solution 2==
 
==Solution 2==
  
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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 1/2 chance of a factor chosen being divisible by 11. The chance of choosing any factor of 11 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>  
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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
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==Solution 3 (generalized)==
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Let <math>N</math> be the positive integer in question. Since <math>N</math> is a multiple of <math>11</math>, we can write <math>N = 11^{e_1}p_1^{e_2}\cdots p_{k-1}^{e_k},</math> the prime factorization of <math>N</math>. To find the total number of divisors divisible by <math>11</math>, observe that there are <math>(e_2 + 1)(e_3 + 1)\cdots (e_k  + 1)</math> divisors not divisible by <math>11</math>. For each power of <math>11</math> greater than <math>1</math>, of which there are <math>e_1</math>, it can be paired with any of these other divisors. Therefore, there are
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<cmath>e_1(e_2+1)(e_3 + 1)\cdots (e_k + 1)</cmath>
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divisors of <math>N</math> that are divisible by <math>11</math>. The total number of divisors of <math>11</math> is <cmath>(e_1 + 1)(e_2 + 1)\cdots (e_k + 1),</cmath> so the probability of choosing a divisor that is divisible by <math>11</math> is
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<cmath>\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}.</cmath>
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For each positive integer less than or equal to <math>100</math>, the highest power of <math>11</math> that divides it is <math>11</math>, so <math>e_1 = 1</math>, meaning that the probability of choosing a divisor (that is divisible by <math>11</math>) of a fixed <math>N</math> is <math>\frac{1}{2}.</math> The probability of choosing any <math>N</math> from the first <math>100</math> positive integers is <math>\frac{1}{100},</math> so the probability of choosing any of these divisors is <math>\frac{1}{100}\cdot \frac{1}{2}.</math> There are <math>9</math> multiples of <math>11</math> less than or equal to <math>100</math>, so the total probability is
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<cmath>9\cdot \frac{1}{100}\cdot \frac{1}{2} = \boxed{\textbf{(B)}\ \frac{9}{200}}.</cmath>
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-Benedict T (countmath1)
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==Video Solution by Power Solve (easy to digest!)==
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https://www.youtube.com/watch?v=jkfsBYzBJbQ
 +
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 +
https://www.youtube.com/watch?v=uaf46N6qP54
 +
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==Video Solution==
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 +
https://youtu.be/O_P3E9hDrYY
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/N2lyYRMuZuk?si=-CyrdswJABoMrnWk&t=825
 +
~Math-X
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2023|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2023|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:57, 6 February 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)

Video Solution by Math-X (First understand the problem!!!)

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

~Math-X


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

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