CGPSC Prelims CSAT 2019 Solved Paper 2

Concept:The unit digit follows a repeating cycle of length $4$ for powers of integers.Explanation:For base $7$, the unit digits cycle as $7, 9, 3, 1$.Since $95 = 4 \times 23 + 3$, the unit digit of $7^{95}$ is the third digit in the cycle, which is $3$.For base $3$, the unit digits cycle as $3, 9, 7, 1$.Since $58 = 4 \times 14 + 2$, the unit digit of $3^{58}$ is the second digit in the cycle, which is $9$.Now, subtract the unit digits: $3 - 9$.Since $3 < 9$, we take a borrow: $13 - 9 = 4$.Therefore, the unit digit of $N$ is $4$.Answer:The digit at the unit place of $N$ is $4$.