Prelims 2026 · CSAT / Quantitative Aptitude · Question 42
The digit in the unit place of the number 6^129 × 7^307 is<br/>
Answer
8
The unit digit of 6^129 is always 6. For 7^n, unit digits cycle as 7, 9, 3, 1 with period 4. Since 307 mod 4 = 3, the unit digit of 7^307 is 3. Therefore the unit digit of 6^129 × 7^307 is 6 × 3 = 18, i.e. 8.