Official Solution: If \(k\) is the number formed by writing the integers from 1 to 999 sequentially (123456789101112...998999), what is the 341st digit in \(k\)? A. 0
B. 1
C. 4
D. 5
E. 9
We begin by determining the number of digits contributed by each group of numbers:
1-digit numbers (1 through 9) contribute 9 digits.
2-digit numbers (10 through 99) contribute 90 * 2 = 180 digits.
3-digit numbers contribute \(3 * x\) digits, where x is the number of 3-digit numbers included.
Setting up the equation to locate the 341st digit:
\(9 + 180 + 3 * x = 341\)
\(x = 50 + \frac{2}{3}\)
This means the 341st digit is in the 51st 3-digit number (since \(50 + \frac{2}{3}\) indicates we need the next number after the first 50).
Starting from 100, the 51st 3-digit number is 100 + 51 - 1 = 150. Since we have \(\frac{2}{3}\) remaining, the required digit is the second digit of 150, which is 5.
Therefore, the 341st digit is 5.
Answer: D