2022 AMC 12B Problem 3

Question

How many of the first ten numbers of the sequence

121,

11211,

1112111,

\ldots

are prime numbers?

0

1

2

3

4

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.

Problem 3 in other years