These questions are based on the following information:\n\nGiven $N = 1! + 2! + 3! + ........ + 99! + 100!$.\n\nFind the last two digits of $N$.

### Concept & Logic To find the last two digits of a number, we must find its remainder when divided by 100. In a factorial series, factorials quickly become multiples of 100 as they accumulate prime factors of 2 and 5. $$N \pmod{100}$$ ### Step-by-Step Solution * **Given:** $N = 1! + 2! + 3! + ........ + 99! + 100!$. (Note: The original text utilizes classical Indian L-bracket notation for factorials, universally translated to modern notation here). * **Calculation:** A factorial $n!$ ends in at least two zeros (meaning it is perfectly divisible by 100) if it contains at least two 5s and two 2s in its prime factorization. * Let's look at $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$. This contains a 5 and a 10 (which provides another 5). Thus, it has two 5s. For any $n \ge 10$, $n! \equiv 0 \pmod{100}$. * Therefore, we only need to calculate the sum of the factorials from $1!$ to $9!$ and find their remainder modulo 100. * Let's evaluate them modulo 100: * $1! = 1$ * $2! = 2$ * $3! = 6$ * $4! = 24$ * $5! = 120 \implies 20$ * $6! = 720 \implies 20$ * $7! = 5040 \implies 40$ * $8! = 40320 \implies 20$ * $9! = 362880 \implies 80$ * Sum these last two digits together: $$\text{Sum} = 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 = 213$$ * The last two digits of 213 are 13. ### Exam Strategy & Shortcut Memorize the last two digits of factorials up to $9!$. Actually, you only need to calculate them iteratively on the fly: $5! = 20$. $6! = 20 \times 6 = 120 \rightarrow 20$. $7! = 20 \times 7 = 140 \rightarrow 40$. $8! = 40 \times 8 = 320 \rightarrow 20$. $9! = 20 \times 9 = 180 \rightarrow 80$. Add them up. Once you hit $10!$, every single term contributes $00$, so you can completely ignore the rest of the infinite-looking series. ### Common Pitfall A common mistake is stopping at $5!$, thinking that since it has a zero, all subsequent terms will have the same zero impact. However, $5!$ to $9!$ only have ONE trailing zero. You need TWO trailing zeros to drop the term entirely when calculating the last two digits. ### Final Answer Therefore, the correct answer is **13**.