UPSC Prelims 2025 CSAT, question 48

Solution.

Answer: (d) 9

1. In $123456789101112\dots$, the nine 1-digit numbers ($1$ to $9$) occupy $9 \times 1 = 9$ digits, and the ninety 2-digit numbers ($10$ to $99$) occupy $90 \times 2 = 180$ digits, for a total of $9 + 180 = 189$ digits.
2. The remaining $489 - 189 = 300$ digits come from 3-digit numbers starting at $100$, which corresponds to exactly $\frac{300}{3} = 100$ three-digit numbers ($100$ through $199$).
3. Therefore, the 489th digit is the last digit of $199$, which is $9$.