Question

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

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

$0$

$1$

$2$

$3$

$4$

答案：A

知识点：质数因式分解

难度评级：1140

小提示：

把第 $n$ 项表示成两个重叠的、各由 $n+1$ 个一组成的数之和

Express the $n$ th term as the sum of two overlapping blocks of $n+1$ ones

大提示：

每一项都可以分解为两个大于一的因数。

Each term in the sequence has an obvious factorization pattern

解答：

我们断言这些数都不可能是质数。

第 $n$ 项可写成 $\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$ $\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$ $= (10^n+1)(\sum_{k=0}^{n} 10^k)\text{。}$ 这说明每一项都能写成两个大于 $1$ 的整数的乘积，所以前十项中没有质数。

所以正确答案是 A。

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the $n$ th number is $\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$ $\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$ $= (10^n+1)(\sum_{k=0}^{n} 10^k).$ This shows that the number can be written as the product of two numbers greater than $1,$ so there are no primes.

Thus, the answer is A .

2022 AMC 10B 第 6 题

答案：A

难度评级：1140

小提示：

大提示：

解答：

其他年份的第 6 题