UPSC Prelims 2025 CSAT, question 67

Solution.

Answer: (a) 0

1. For every positive integer $k \ge 1$, $9^k$ is an odd multiple of $3$, so $9^k \equiv 3 \pmod{6}$.
2. The sum $9^3 + 9^4 + \dots + 9^{100}$ contains $100 - 3 + 1 = 98$ terms, each congruent to $3 \pmod{6}$.
3. Since $98$ is even, $98 \times 3 = 294$ is a multiple of $6$, so the remainder is $0$.