The digit in the unit place of the number 6¹²⁹ × 7³⁰⁷ is

Answer: (c) 8

Any power of 6 ends in 6. Powers of 7 end in 7, 9, 3, 1 in a cycle of four; 307 leaves remainder 3 on division by 4, so 7³⁰⁷ ends in 3. The product ends in the last digit of 6 × 3 = 18, i.e., 8.

1. 6¹²⁹ ends in 6.
2. 7¹, 7², 7³, 7⁴ end in 7, 9, 3, 1; the cycle length is 4.
3. 307 = 4 × 76 + 3, so 7³⁰⁷ ends in 3.
4. 6 × 3 = 18, so the unit digit is 8.

Remember · Unit digits: 0, 1, 5, 6 never change; 2, 3, 7, 8 cycle every 4 — use the exponent's remainder on division by 4.

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