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

(Solution)
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It follows that the first ten terms are divisible by <math>11,101,1001,\ldots,</math> respectively.
 
It follows that the first ten terms are divisible by <math>11,101,1001,\ldots,</math> respectively.
  
Therefore, there are \boxed{\textbf{(A) } 0} prime numbers in this sequence.
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Therefore, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 15:32, 17 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

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 first ten terms are divisible by $11,101,1001,\ldots,$ respectively.

Therefore, there are $\boxed{\textbf{(A) } 0}$ prime numbers in this sequence.

~MRENTHUSIASM

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
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

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