1983 AMC 12 Problem 16

Question

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

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

After

189

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

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.

Problem 16 in other years