2022 AMC 10B Problems/Problem 6

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 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 (If you don't see the above/are running low on time)

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 $\boxed{\textbf{(A)} 0}.$

~Dhillonr25

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2022 AMC 10B Problems/Problem 6

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Problem

Solution

Solution 2 (If you don't see the above/are running low on time)

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