1983 AMC 12 第 16 题

Question

令

\begin{aligned} x&=0.123456789101112\\ &\phantom{={}}\ldots998999 \end{aligned}\text{，}

其中小数部分的数字是将整数

1

到

999

依次写下而得到的。小数点右侧第

1983

位数字是

Let

\begin{aligned} x&=0.123456789101112\\ &\phantom{={}}\ldots998999, \end{aligned}

where the digits are obtained by writing the integers

1

through

999

in order. The

1983

rd digit to the right of the decimal point is

2

3

5

7

8

先数一位数和两位数共提供多少位数字

First count the digits contributed by the one- and two-digit integers

经过前

189

位后，在三位数中定位剩余的位置

After

189

digits, locate the remaining position within the three-digit integers

一位数提供

9

位，两位数提供

90\cdot2=180

位，共

189

位。因此所求数字是三位数部分中的第

1983-189=1794

位。由于

1794=598\cdot3

，它是第

598

个三位数的末位数字；该三位数为

100+597=697

。所以这位数字是

7

。

所以正确答案是 D。

The one-digit integers contribute

9

digits and the two-digit integers contribute

90\cdot2=180,

for

189

total. Thus the desired digit is the

1983-189=1794

th digit among the three-digit integers. Since

1794=598\cdot3,

it is the last digit of the

598

th three-digit integer, namely

100+597=697.

That digit is

7.

Therefore, the correct answer is D.

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