Question

Consider a sequence formed by concatenating the positive integers in increasing order: 12345678910111213... Which among the following is the closest to the 2028th 4th digit in this infinite sequence?

• A. 4
• B. 7
• C. 1
• D. 9

Hint:

When dealing with large concatenated sequences, break down the sequence into groups (e.g., one-digit numbers, two-digit numbers) and calculate the contribution of each group to locate the desired digit.

Answer 1

The Correct Option is C

We are asked to find the 2028th 4th digit in the sequence formed by concatenating the integers 1, 2, 3, 4, ..., in order. The goal is to figure out which digit is at the 2028th 4th place. The digits from the numbers 1 to 9 contribute 9 digits in total. From 10 to 99, there are 90 two-digit numbers, contributing 180 digits. From 100 to 999, there are 900 three-digit numbers, contributing 2700 digits. 

So, to find the 2028th 4th digit, we look at how many digits are contributed by the two-digit and three-digit numbers: 

- The first 189 digits come from the numbers 1 to 99. 

- The next 2028 

- 189 = 1839 digits are needed, which come from the three-digit numbers. 

Since each three-digit number contributes 3 digits, we divide 1839 by 3: \[ 1839 \div 3 = 613 \] Thus, the 2028th digit is the third digit of the 613th three-digit number. The 613th three-digit number is: \[ 100 + 613 - 1 = 712 \] The third digit of 712 is \( 1 \). Thus, the 2028th 4th digit in the sequence is \( 1 \).