Difference between revisions of "2022 AMC 10B Problems/Problem 6"

Revision as of 03:19, 7 February 2023

The following problem is from both the 2022 AMC 10B #6 and 2022 AMC 12B #3, so both problems redirect to this page.

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

~ab2024

Solution 3 (Educated Guess)

Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$ . Because this problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{\textbf{(A) } 0}.$

~Dhillonr25

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 39: / Line 39: @@
 ~Dhillonr25
-==Solution 4 (Detailed Explanation of Solution 1)==
-Denote this sequence as <math>a_{n}</math>, then we can find that
-<cmath>\begin{align*}
-a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\
-a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\
-a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\
-& \ \vdots
-\end{align*}</cmath>
-Hence we can induct the general term is
-<cmath>\begin{align*}
-a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\
-&= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\
-&= \left(10^n+1\right)\sum_{k=0}^{n}10^k.
-\end{align*}</cmath>
-Hence, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence.
-~PythZhou
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