2022 AMC 10B Problem 6

Testing big numbers like $1112111$ for primality by hand is hopeless, so look for a pattern in how the terms are built.

1. Try to factor the first few terms.
$$121 = 11 \times 11, \qquad 11211 = 111 \times 101, \qquad 1112111 = 1111 \times 1001.$$
Each term is "a number made of ones" times "a $1$, some zeros, and a $1$".
2. See why the multiplication works. Take $111 \times 101$. Multiplying by $101$ means adding $111 \times 100$ and $111 \times 1$:
$$\begin{array}{r} 11100 \\ +\ \ \ \ 111 \\ \hline 11211 \end{array}$$
The two copies of $111$ are shifted and overlap in just one column, where $1 + 1 = 2$. Every other column has a single $1$.
3. It works for every term. The term with a block of ones on each side of the $2$ equals (one more $1$ than that block) times ($1$, then zeros, then $1$). For example, the 4th term $111121111 = 11111 \times 10001$. The same overlapping-by-one-column addition gives it every time.
4. So no term is prime. Both factors are bigger than $1$ in every case, so each term is a product of two smaller whole numbers. None of the first ten terms is prime.

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

A number described by its digit pattern usually has a simple multiplication hiding behind it. Here, the single $2$ in the middle is the mark left when two shifted copies of $111\ldots1$ are added and overlap in one column. Factoring the first two or three terms and then explaining the pattern is much faster and safer than testing big numbers for primes.

With algebra: let $a = 10^n$, where $n$ is the number of ones on each side of the $2$. The term equals $111\ldots1$ (with $2n+1$ ones) plus $a$, because adding $a$ turns the middle $1$ into a $2$. The number with $2n+1$ ones is $\frac{10a^2 - 1}{9}$, so the term is
$$\frac{10a^2 - 1}{9} + a = \frac{10a^2 + 9a - 1}{9} = \frac{(10a - 1)(a + 1)}{9}.$$
Here $\frac{10a - 1}{9}$ is the number made of $n + 1$ ones and $a + 1 = 10^n + 1$, which is the same factoring as above.

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 $\boxed{\textbf{(A)}\ 0}$.