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Bunuel, twobagels

1* 9 + 2 * 90 + 3 * x = 1983
x = 598

=> 598 three digit sets starting at 100
The answer to this question remains 7, but if we were asked the 1981st digit would be be 6 or 5 ?
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shaurya_gmat

Let x = 0.12345678910111213. . . 998999, where the digits are obtained by writing the integers 1 through 999 in that order. The 1983rd digit to the right of the decimal point is?

A. 2
B. 3
C. 5
D. 7
E. 8

Bunuel, twobagels

1* 9 + 2 * 90 + 3 * x = 1983
x = 598

=> 598 three digit sets starting at 100
The answer to this question remains 7, but if we were asked the 1981st digit would be be 6 or 5 ?

x = 0.12345678910111213... 696697...998999

In the sequence, 7 is the 1983rd digit, so 6 will be the 1981st digit.

We start by solving for the number of digits:

  • 1-digit numbers contribute 9 digits: 1 through 9.
  • 2-digit numbers contribute 2 * 90 = 180 digits: 10 through 99.
  • 3-digit numbers contribute 3 * x digits, where x is the number of 3-digit numbers.

Setting up the equation:

9 + 2 * 90 + 3 * x = 1981
9 + 180 + 3 * x = 1981
\(x = 597 \frac{1}{3}\)

Starting from 100, the 597th 3-digit number is 100 + 597 - 1 = 696. Since we have 1/3 left, we need the first digit of the next number, which is 6. Therefore, the 1981st digit is 6.
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