2022 AMC 12B 第 3 题

Look for a factorization of each term into two integers both greater than $1$

Try writing a term as a repunit times a number of the form $10^n + 1,$ for example $11211 = 111 \cdot 101$

The $n$th term consists of $n$ ones, then a $2,$ then $n$ ones. It factors as a repunit times a number of the form $10^n + 1:$ $$\begin{aligned} &121 = 11 \cdot 11, \\ &\quad 11211 = 111 \cdot 101, \\ &\quad 1112111 = 1111 \cdot 1001, \end{aligned}$$ and in general the $n$th term equals $\underbrace{1\cdots1}_{n+1} \cdot (10^n + 1).$

For every $n \ge 1$ both factors exceed $1,$ so every term is composite. None of the ten numbers is prime.

Thus, the correct answer is A.