2019 AIME I 第 1 题

Question

考虑整数

\begin{aligned} &N = 9 + 99 + 999 + 9999 \\ &\quad {}+ \cdots + \underbrace{99\ldots99}_{\text{321 位数字}} \end{aligned}\text{。}

求

N

的各位数字之和。

Consider the integer

\begin{aligned} &N = 9 + 99 + 999 + 9999 \\ &\quad {}+ \cdots + \underbrace{99\ldots99}_{\text{321 digits}}. \end{aligned}

Find the sum of the digits of

N.

答案

知识点：数字位值

每一项都是

10^k - 1

，所以

N

是一长串一（后面接一个零）再减去

321

Each summand is

10^k - 1,

so

N

is a long string of ones (followed by a zero) minus

321

先看减数

321

：用

11\ldots110

减去它只会影响最后四位；数一数剩下多少个一

Subtracting

321

from

11\ldots110

disturbs only the last four digits; count the ones that survive

每一项都是

10^k - 1

，所以

\begin{aligned} N &= \sum_{k=1}^{321} \left(10^k - 1\right) \\ &= \underbrace{11\ldots1}_{321}0 - 321 \end{aligned}\text{。}

这个减法只改变最后四位：

1110 - 321 = 789

，所以最后四位变为

0789

。因此

N

由

318

个一后接

0789

组成，数字和为

318 + 0 + 7 + 8 + 9 = 342

。

Each summand is

10^k - 1,

so

\begin{aligned} N &= \sum_{k=1}^{321} \left(10^k - 1\right) \\ &= \underbrace{11\ldots1}_{321}0 - 321. \end{aligned}

The subtraction changes only the last four digits:

1110 - 321 = 789,

so those four digits become

0789.

Thus

N

consists of

318

ones followed by

0789,

and the digit sum is

318 + 0 + 7 + 8 + 9 = 342.

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