Track / Stage 3 / 40 of 260 #40 of 1964

Problem 40

AMC 10/12, early questions

Number theory Difficulty 3.1 Multiple choice

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

Pick one

Official solution

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

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.