Maths Olympiad Prep

Track / Stage 3 / 36 of 260 #36 of 1964

Problem 36

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,…121, 11211, 1112111, \ldots are prime numbers?

Pick one

Official solution

The nnth term of this sequence is
∑k=n2n10k+∑k=0n10k=10n∑k=0n10k+∑k=0n10k=(10n+1)∑k=0n10k.\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
121=11⋅11,11211=101⋅111,1112111=1001⋅1111, ⋮\begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*}
Therefore, there are (A) 0\boxed{\textbf{(A) } 0} prime numbers in this sequence.
~MRENTHUSIASM

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.