What is the unit digit in the expansion of (57242)^(9×7×5×3×1)?

Answer: (a) 2

Only the last digit of the base, 2, matters, and the unit digits of powers of 2 repeat every 4. The exponent 9 × 7 × 5 × 3 × 1 = 945 leaves remainder 1 on division by 4, so the unit digit is that of 2¹, which is 2.

1. The base ends in 2 and the exponent is 9 × 7 × 5 × 3 × 1 = 945, so we need the unit digit of 2^945.
2. Unit digits of powers of 2 run 2, 4, 8, 6 and then repeat (cycle of 4).
3. 945 ÷ 4 leaves remainder 1.
4. So the unit digit is the first in the cycle: 2.

Remember · Unit digit of a power: keep the base's last digit, find its cycle (2, 3, 7, 8 repeat every 4), use exponent mod 4.

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