For an integer $n$, $n! = n(n - 1) (n - 2)............. 3.2.1.$ Then, $1! + 2! + 3! + ........ + 100!$ when divided by $5$ leaves remainder

### Concept & Logic In modular arithmetic, the remainder of a sum is equal to the sum of the remainders. For any factorial $n!$ where $n \ge 5$, the expansion includes the factor $5$ (e.g., $5! = 5 \times 4 \times 3 \times 2 \times 1$). Therefore, all factorials from $5!$ onwards are perfectly divisible by $5$ and leave a remainder of $0$. ### Step-by-Step Solution * **Given:** * Expression: $1! + 2! + 3! + ... + 100!$ * Divisor: $5$ * **Calculation / Deduction:** 1. Break the series into two parts: terms before $5!$ and terms from $5!$ onwards. $$ (1! + 2! + 3! + 4!) + (5! + 6! + ... + 100!) $$ 2. Analyze the second part. Since $5!, 6!, 7!$, etc., all contain $5$ as a multiplier, their remainders when divided by $5$ are strictly $0$. 3. Calculate the exact value of the first part: $1! = 1$ $2! = 2$ $3! = 6$ $4! = 24$ 4. Sum these initial terms: $1 + 2 + 6 + 24 = 33$ 5. Find the remainder when $33$ is divided by $5$. $33 / 5 = 6$ with a remainder of $3$. ### Exam Strategy & Shortcut Recognize immediately that $5!$ and beyond contribute nothing to the remainder when dividing by $5$. Simply compute $1 + 2 + 6 + 24 = 33$, and visually see that $33$ ends in $3$. Since numbers ending in $0$ or $5$ are divisible by $5$, the remainder is just the units digit $3$. ### Common Pitfall Students often get intimidated by the $100!$ and attempt to find a complex pattern or formula for the entire series, wasting time. Remembering that factorials quickly absorb all small prime factors is key. ### Final Answer Therefore, the correct answer is 3.