Question

Travis 要照看难缠的 Thompson 三胞胎。知道他们喜欢大数，Travis 为他们设计了一个数数游戏。先由 Tadd 说数字 $1$ 然后 Todd 必须说接下来的两个数（ $2$ 和 $3$ ），接着 Tucker 必须说接下来的三个数（ $4$ $5$ $6$ ），再由 Tadd 说接下来的四个数（ $7$ $8$ $9$ $10$ ）。此后仍按三个孩子的顺序轮流，每个孩子说的数都比前一个孩子多一个，直到数到 $10,000$ 。Tadd 说出的第 $2019$ 个数是多少？

Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1,$ then Todd must say the next two numbers ( $2$ and $3$ ), then Tucker must say the next three numbers ( $4,$ $5,$ $6$ ), then Tadd must say the next four numbers ( $7,$ $8,$ $9,$ $10$ ), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$ th number said by Tadd?

$5743$

$5885$

$5979$

$6001$

$6011$

答案：C

知识点：等差数列三角形数求和

难度评级：2080

解答：

Tadd 每次发言的长度依次为 $1,4,7,\ldots$ 。在 Tadd 发言 $n$ 次后，他共说了个数。 $\sum_{i=1}^n (3i-2)=\frac{3n^2-n}{2}$

当 $n=36$ 时，这个总数为 $1926$ ；当 $n=37$ 时则为 $2035$ 。因此，Tadd 说出的第 $2019$ 个数，是他第 $37$ 次发言中的第 $(2019-1926)=93$ 个数。

在这次发言之前，三个孩子已经完成了长度从 $1$ 到 $108$ 的各轮发言，共说了 $1+2+\cdots+108=5886$ 个数。下一轮的第 $93$ 个数是 $5886+93=5979$ 。所以正确答案是 C。

Tadd speaks on turns of lengths $1,4,7,\ldots$ . After $n$ of Tadd's turns, he has said $\sum_{i=1}^n (3i-2)=\frac{3n^2-n}{2}$ numbers.

For $n=36$ , this total is $1926$ , while for $n=37$ it is $2035$ . Therefore Tadd's $2019$ th number is the $(2019-1926)=93$ rd number of his $37$ th turn.

Before that turn, the children have completed turns of lengths $1$ through $108$ , saying $1+2+\cdots+108=5886$ numbers. The $93$ rd number of the next turn is $5886+93=5979$ . Thus, C is the correct answer.

2019 AMC 10A 第 23 题

答案：C

难度评级：2080

解答：

其他年份的第 23 题