AMC10 2022 B
AMC10 2022 B · Q6
AMC10 2022 B · Q6. 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 $121, 11211, 1112111, \ldots$ 的前十项中有多少项是质数？
(A)
0
0
(B)
1
1
(C)
2
2
(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
\[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\]
因此各项为
\begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*}
因此，该数列中质数的个数为 $\boxed{\textbf{(A) } 0}$。
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