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Difference between revisions of "2022 AMC 10B Problems/Problem 6"

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{{duplicate|[[2022 AMC 10B Problems/Problem 6|2022 AMC 10B #6]] and [[2022 AMC 12B Problems/Problem 3|2022 AMC 12B #3]]}}
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==Problem==
 
==Problem==
 
How many of the first ten numbers of the sequence <math>121, 11211, 1112111, \ldots</math> are prime numbers?
 
How many of the first ten numbers of the sequence <math>121, 11211, 1112111, \ldots</math> are prime numbers?
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<math>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4</math>
 
<math>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4</math>
  
−
==Solution==
+
==Solution 1 (Generalization)==
 
The <math>n</math>th term of this sequence is
 
The <math>n</math>th term of this sequence is
 +
<cmath>\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.</cmath>
 +
It follows that the terms are
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\left(10^{2n}+10^{2n-1}+\cdots+10^{n}\right)+\left(10^{n}+10^{n-1}+\cdots+10^{0}\right) &= 10^n\left(10^{n}+10^{n-1}+\cdots+10^{0}\right)+\left(10^{n}+10^{n-1}+\cdots+10^{0}\right) \\
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121 &= 11\cdot11, \\
−
&= \left(10^n + 1\right)\left(10^{n}+10^{n-1}+\cdots+10^{0}\right).
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11211 &= 101\cdot111, \\
 +
1112111 &= 1001\cdot1111, \\
 +
& \ \vdots
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
It follows that the first ten terms are divisible by <math>11,101,1001,\ldots,</math> respectively.
+
Therefore, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence.
 +
 
 +
~MRENTHUSIASM
  
 +
==Solution 2 (Detailed Explanation of Solution 1)==
 +
 +
Denote this sequence as <math>a_{n}</math>, then we can find that
 +
<cmath>\begin{align*}
 +
a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\
 +
a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\
 +
a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\
 +
& \ \vdots
 +
\end{align*}</cmath>
 +
So, we can induct that the general term is
 +
<cmath>\begin{align*}
 +
a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\
 +
&= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\
 +
&= \left(10^n+1\right)\sum_{k=0}^{n}10^k.
 +
\end{align*}</cmath>
 
Therefore, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence.
 
Therefore, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence.
  
−
~MRENTHUSIASM
+
~PythZhou
 +
 
 +
==Solution 3 (Simple Sums)==
 +
 
 +
Observe how
 +
<cmath>\begin{align*}
 +
121 &= 110 + 11, \\
 +
11211 &= 11100 + 111, \\
 +
1112111 &= 1111000 + 1111, \\
 +
& \ \vdots
 +
\end{align*}</cmath>
 +
all take the form of <cmath>\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).</cmath>
 +
Factoring each of the sums, we have <cmath>11(10+1), 111(100+1), 1111(1000+1), \ldots</cmath> respectively. With each number factored, there are <math>\boxed{\textbf{(A) } 0}</math> primes in the set.
 +
 
 +
~ab2024
 +
 
 +
==Solution 4 (Educated Guess)==
 +
 
 +
Note that <math>121</math> is divisible by <math>11</math> and <math>11211</math> is divisible by <math>3</math>. Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is <math>\boxed{\textbf{(A) } 0}.</math>
 +
 
 +
~Dhillonr25
 +
 
 +
==Video Solution(1-16)==
 +
https://youtu.be/SCwQ9jUfr0g
 +
 
 +
~~Hayabusa1
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/_KNR0JV5rdI?t=562
 +
 
 +
==Video Solution by Math4All999==
 +
https://youtu.be/5QYh3hNaDa0?feature=shared
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2022|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2022|ab=B|num-b=5|num-a=7}}
 +
{{AMC12 box|year=2022|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:18, 20 March 2024

The following problem is from both the 2022 AMC 10B #6 and 2022 AMC 12B #3, so both problems redirect to this page.

Problem

How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?

$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$

Solution 1 (Generalization)

The $n$th term of this sequence is \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] It follows that the terms are \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} Therefore, there are $\boxed{\textbf{(A) } 0}$ prime numbers in this sequence.

~MRENTHUSIASM

Solution 2 (Detailed Explanation of Solution 1)

Denote this sequence as $a_{n}$, then we can find that \begin{align*} a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ & \ \vdots \end{align*} So, we can induct that the general term is \begin{align*} a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= \left(10^n+1\right)\sum_{k=0}^{n}10^k. \end{align*} Therefore, there are $\boxed{\textbf{(A) } 0}$ prime numbers in this sequence.

~PythZhou

Solution 3 (Simple Sums)

Observe how \begin{align*} 121 &= 110 + 11, \\ 11211 &= 11100 + 111, \\ 1112111 &= 1111000 + 1111, \\ & \ \vdots \end{align*} all take the form of \[\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).\] Factoring each of the sums, we have \[11(10+1), 111(100+1), 1111(1000+1), \ldots\] respectively. With each number factored, there are $\boxed{\textbf{(A) } 0}$ primes in the set.

~ab2024

Solution 4 (Educated Guess)

Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$. Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{\textbf{(A) } 0}.$

~Dhillonr25

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI?t=562

Video Solution by Math4All999

https://youtu.be/5QYh3hNaDa0?feature=shared

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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
2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions

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