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

(→‎Solution 1)
(→‎Video Solution by Math-X (First fully understand the problem!!!))
 
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In order for the divisor chosen to be a multiple of <math>11</math>, the original number chosen must also be a multiple of <math>11</math>. Among the first <math>100</math> 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 <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{\textbf{(B)}~\frac{9}{200}}.</math>
 
In order for the divisor chosen to be a multiple of <math>11</math>, the original number chosen must also be a multiple of <math>11</math>. Among the first <math>100</math> 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 <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{\textbf{(B)}~\frac{9}{200}}.</math>
  
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<math>11 = 11 - \frac{1}{2}\\
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<math>11 = 1, 11 \Rightarrow \frac{1}{2}\\
−
22 = 2 \times 11: 11, 22 - \frac{1}{2}\\
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22 = 2 \times 11: 1, 2, 11, 22 \Rightarrow \frac{1}{2}\\
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33 = 3 \times 11: 11, 33 - \frac{1}{2}\\
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33 = 3 \times 11: 1, 3, 11, 33 \Rightarrow \frac{1}{2}\\
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44 = 2^2 \times 11: 11, 22, 44 - \frac{1}{2}\\
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44 = 2^2 \times 11: 1, 2, 4, 11, 22, 44 \Rightarrow \frac{1}{2}\\
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55 = 5 \times 11: 11, 55 - \frac{1}{2}\\
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55 = 5 \times 11: 1, 5, 11, 55 \Rightarrow \frac{1}{2}\\
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66 = 2 \times 3 \times 11: 11, 22, 33, 66 - \frac{1}{2}\\
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66 = 2 \times 3 \times 11: 1, 2, 3, 6, 11, 22, 33, 66 \Rightarrow \frac{1}{2}\\
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77 = 7 \times 11: 11, 77 - \frac{1}{2}\\
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77 = 7 \times 11: 1, 7, 11, 77 \Rightarrow \frac{1}{2}\\
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88 = 2^3 \times 11: 11, 22, 44, 88 - \frac{1}{2}\\
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88 = 2^3 \times 11: 1, 2, 4, 8, 11, 22, 44, 88 \Rightarrow \frac{1}{2}\\
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99 = 3^2 \times 11: 11, 33, 99 - \frac{1}{2}</math>
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99 = 3^2 \times 11: 1, 3, 9, 11, 33, 99 \Rightarrow \frac{1}{2}</math>
  
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~vaisri ~walmartbrian ~Shontai ~(minor formatting changes by Marshall_Huang)
+
~vaisri  
 +
 
 +
~walmartbrian  
 +
 
 +
~Shontai  
 +
 
 +
~(minor formatting changes by Marshall_Huang)
 +
 
 +
~Formatting changes to avoid confusion by SHREYU.MEDATATI
  
 
==Solution 2==
 
==Solution 2==
  
−
There are <math>\text{floor}(100/11) = 9</math> multiples of <math>11</math> under <math>100</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>  
+
There are <math>\left\lfloor\frac{100}{11}\right\rfloor = 9</math> multiples of <math>11</math> under <math>100</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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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  
 
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  
 
<cmath>e_1(e_2+1)(e_3 + 1)\cdots (e_k + 1)</cmath>
 
<cmath>e_1(e_2+1)(e_3 + 1)\cdots (e_k + 1)</cmath>
−
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  
+
divisors of <math>N</math> that are divisible by <math>11</math>. The total number of divisors of <math>N</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  
 
<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>
 
<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>
 
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
 
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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-Benedict T (countmath1)
 
-Benedict T (countmath1)
 +
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=BATFbSs7VssVJpvm&t=3050
 +
~little-fermat
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=J8YkZnOt7AoOgUcS&t=3530
 +
~Math-X
 +
 +
==Video Solution ⚡️ 1 min solution ⚡️ ==
 +
 +
https://youtu.be/CP0Arc_82Y8
 +
 +
<i> ~Education, the Study of Everything </i>
  
 
==Video Solutions==
 
==Video Solutions==
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https://youtu.be/N2lyYRMuZuk?si=-CyrdswJABoMrnWk&t=825
 
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~Math-X
 
  
 
==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}}

Latest revision as of 06:30, 5 November 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 $\boxed{\textbf{(B)}~\frac{9}{200}}.$

$11 = 1, 11 \Rightarrow \frac{1}{2}\\ 22 = 2 \times 11: 1, 2, 11, 22 \Rightarrow \frac{1}{2}\\ 33 = 3 \times 11: 1, 3, 11, 33 \Rightarrow \frac{1}{2}\\ 44 = 2^2 \times 11: 1, 2, 4, 11, 22, 44 \Rightarrow \frac{1}{2}\\ 55 = 5 \times 11: 1, 5, 11, 55 \Rightarrow \frac{1}{2}\\ 66 = 2 \times 3 \times 11: 1, 2, 3, 6, 11, 22, 33, 66 \Rightarrow \frac{1}{2}\\ 77 = 7 \times 11: 1, 7, 11, 77 \Rightarrow \frac{1}{2}\\ 88 = 2^3 \times 11: 1, 2, 4, 8, 11, 22, 44, 88 \Rightarrow \frac{1}{2}\\ 99 = 3^2 \times 11: 1, 3, 9, 11, 33, 99 \Rightarrow \frac{1}{2}$

~vaisri

~walmartbrian

~Shontai

~(minor formatting changes by Marshall_Huang)

~Formatting changes to avoid confusion by SHREYU.MEDATATI

Solution 2

There are $\left\lfloor\frac{100}{11}\right\rfloor = 9$ multiples of $11$ under $100$. 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 $N$ 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 Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=BATFbSs7VssVJpvm&t=3050 ~little-fermat

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

https://youtu.be/GP-DYudh5qU?si=J8YkZnOt7AoOgUcS&t=3530 ~Math-X

Video Solution ⚡️ 1 min solution ⚡️

https://youtu.be/CP0Arc_82Y8

~Education, the Study of Everything

Video Solutions

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


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

Video Solution

https://youtu.be/O_P3E9hDrYY

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


See Also

2023 AMC 10A (Problems • Answer Key • Resources)
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Problem 13 		Followed by
Problem 15
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