AMC12 2022 B

AMC12 2022 B · Q3

AMC12 2022 B · Q3. It mainly tests Divisibility & factors, Primes & prime factorization.

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

数列 $121, 11211, 1112111, \ldots$ 的前十项中有多少是质数？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (A)

正确答案：（A）

Solution

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.

该数列的第 $n$ 项为 \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] 从而各项为 \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} 因此，该数列中没有质数。\boxed{\textbf{(A) } 0}

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