UPSC NDA I 2023 Mathematics Solved Paper

Concept:The remainder when $S$ is divided by $8$ is found by summing only the terms not divisible by $8$.Explanation:$S = 0! + 1! + 2! + 3! + 4! + \dots + 100!$.For $n \ge 4$, $n!$ contains the factor $2^3 = 8$ because $4! = 24 = 8 \times 3$ and higher factorials include $4!$ as a factor.Thus, all terms from $4!$ onward are multiples of $8$ and contribute $0$ remainder.Only $0! + 1! + 2! + 3!$ need to be considered.Calculate: $0! = 1$, $1! = 1$, $2! = 2$, $3! = 6$. Their sum is $1 + 1 + 2 + 6 = 10$.Now find remainder of $10$ modulo $8$: $10 = 8 \times 1 + 2$, so remainder is $2$.Hence, $S \equiv 2 \pmod{8}$.Answer:2 (Option C)