What is the remainder when 9³ + 9⁴ + 9⁵ + 9⁶ + ... + 9¹⁰⁰ is divided by 6 ?
- (a) 0 ✓ UPSC's answer
- (b) 1
- (c) 2
- (d) 3
Why the answer is (a)
• 9 ≡ 3 (mod 6), and 3² = 9 ≡ 3 (mod 6), so every positive power 9ⁿ ≡ 3 (mod 6).
• The series 9³ + 9⁴ + ... + 9¹⁰⁰ contains 100 − 3 + 1 = 98 terms.
• Each of the 98 terms leaves remainder 3, so the sum leaves remainder 98 × 3 = 294 (mod 6).
• 294 ÷ 6 = 49 with remainder 0, so the whole sum is divisible by 6.
• Therefore the required remainder is 0, which is option (a).
Why the other options are wrong
- (b) 1
- The sum leaves remainder 294 modulo 6, which is 0, not 1.
- (c) 2
- The sum leaves remainder 294 modulo 6, which is 0, not 2.
- (d) 3
- The sum leaves remainder 294 modulo 6, which is 0, not 3.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.