2022 AMC 10B Problem 6

The $n$th term consists of $n$ ones, a $2$, and $n$ more ones. Try to factor the first few:
$$121 = 11 \cdot 11,\qquad 11211 = 111 \cdot 101,\qquad 1112111 = 1111 \cdot 1001.$$

This pattern always holds. Let $R$ be the number made of $n+1$ ones. Then
$$R \cdot (10^n + 1) = R \cdot 10^n + R.$$
The first summand is $n+1$ ones followed by $n$ zeros; the second is $n+1$ ones. When added, they overlap in exactly one digit position (the $(n+1)$st from the right), which becomes $1 + 1 = 2$, and every other position gets a single $1$. The result is $n$ ones, a $2$, and $n$ ones: precisely the $n$th term.

Both factors $R$ and $10^n + 1$ exceed $1$, so every term is composite. None of the first ten (or of any) terms is prime.

The answer is $\boxed{\textbf{(A)}\ 0}$.

Numbers described by their digit pattern usually have a hidden algebraic form; here the "2 in the middle" is the fingerprint of two shifted copies of a repunit overlapping in one digit. Verifying the factorization on the first couple of terms and then explaining it in general is far more reliable than trying to test large numbers for primality.

Algebraically, with $a = 10^n$ the term is $\frac{10a^2 - 1}{9} + a = \frac{10a^2 + 9a - 1}{9} = \frac{(10a-1)(a+1)}{9}$, and $\frac{10a-1}{9}$ is the repunit with $n+1$ ones.

• Testing only the first term (121 = 11²), seeing that the others are not obviously composite, and guessing that one or two of them are prime.
• Checking divisibility by $3$ via digit sums (the digit sum $2n+2$ is often not a multiple of $3$) and concluding a term is prime when the test fails.