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Difference between revisions of "2022 AMC 10B Problems/Problem 6"
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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>
  
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==Solution==
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==Solution 1 (Generalization)==
 
The <math>n</math>th term of this sequence is
 
The <math>n</math>th term of this sequence is
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2 (If you don't see the above/are running low on time)==
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==Solution 2 (Educated Guesses)==
  
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Note that it's obvious that 121 is divisible by 11 and 11211 is divisible by 3; therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two digit prime divisibility or use obscure theorems; therefore, the answer is <math>\boxed{\textbf{(A)} 0}.</math>
+
Note that it's obvious that <math>121</math> is divisible by <math>11</math> and <math>11211</math> is divisible by <math>3;</math> therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two-digit prime divisibility or use obscure theorems. So, the answer is <math>\boxed{\textbf{(A) } 0}.</math>
  
 
~Dhillonr25
 
~Dhillonr25

Revision as of 17:59, 30 November 2022

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 \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) \\ &= \left(10^n + 1\right)\left(10^{n}+10^{n-1}+\cdots+10^{0}\right). \end{align*} 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 (Educated Guesses)

Note that it's obvious that $121$ is divisible by $11$ and $11211$ is divisible by $3;$ therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two-digit prime divisibility or use obscure theorems. So, the answer is $\boxed{\textbf{(A) } 0}.$

~Dhillonr25

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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