What is the remainder when 9³ + 9⁴ + 9⁵ + 9⁶ + ... + 9¹⁰⁰ is divided by 6?

Answer: (a) 0

Every power of 9 leaves remainder 3 when divided by 6, and there are 98 terms. 98 × 3 = 294 is a multiple of 6, so the remainder is 0.

1. 9 leaves remainder 3 when divided by 6, and 3 × 3 = 9 again leaves 3. So every power 9ᵏ leaves remainder 3.
2. Number of terms from 9³ to 9¹⁰⁰: 100 − 3 + 1 = 98.
3. Sum of remainders = 98 × 3 = 294 = 6 × 49, so the remainder is 0.
4. Check: each term is odd and a multiple of 3; 98 odd terms add to an even number, and an even multiple of 3 is divisible by 6.

Remember · For the remainder of a sum, reduce each term first; count terms carefully (3 to 100 is 98 terms).

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). · Report an error