A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13?

Answer: (a) 11

999999 (six 9s) = 999 × 1001 is divisible by 13, so each block of six 9s leaves nothing over. 99 nines are 16 such blocks followed by 999, so N leaves the same remainder as 999, which is 11.

1. 999999 = 999 × 1001 = 999 × 7 × 11 × 13, so it is divisible by 13.
2. 99 = 6 × 16 + 3, so N is sixteen blocks of 999999 followed by 999.
3. The first 96 nines form a multiple of 999999 (and so of 13), shifted three places; hence N leaves the same remainder as 999.
4. 999 = 13 × 76 + 11, so the remainder is 11.

Remember · Six identical digits (aaaaaa) are always divisible by 7, 11 and 13; remove blocks of six and work with the digits left over.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). · Report an error