2022 AMC 10B Problems/Problem 6

Problem

How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?

$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$

Solution 1 (Generalization)

The $n$ th term of this sequence is $\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.$ It follows that the terms are $\begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*}$ Therefore, there are $\boxed{\textbf{(A) } 0}$ prime numbers in this sequence.

~MRENTHUSIASM

Solution 2 (Simple Sums)

Observe how $121 = 110+11$ and $11211 = 11100 + 111$ and $1112111 =1111000 + 1111$ all take the form of $\overbrace{111\ldots}^{n+1}\overbrace{00\ldots}^{n} + \overbrace{111\ldots}^{n+1} = \overbrace{111\ldots}^{n+1}(10^{n} + 1).$ Factoring each of the sums, we have $11(10+1), 111(100+1),$ and $1111(1000+1)$ respectively. With each number factored, there are $\boxed{\textbf{(A) } 0}$ primes in the set.

~ab2024

Solution 3 (Educated Guesses)

Note that it's obvious that $121$ is divisible by $11$ and $11211$ is divisible by $3;$ therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two-digit prime divisibility or use obscure theorems. So, the answer is $\boxed{\textbf{(A) } 0}.$

~Dhillonr25

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2022 AMC 10B Problems/Problem 6

Contents

Problem

Solution 1 (Generalization)

Solution 2 (Simple Sums)

Solution 3 (Educated Guesses)

See Also