abdulrafay

ou're looking for the remainder when the sum of factorials from 1! to 100! is divided by 2.

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

...

Notice that starting from 2!, every factorial is an even number (because it includes a factor of 2).

So, when we divide each factorial by 2:

1! / 2 leaves a remainder of 1.

2! / 2 leaves a remainder of 0.

3! / 2 leaves a remainder of 0.

4! / 2 leaves a remainder of 0.

...

100! / 2 leaves a remainder of 0.

Therefore, the sum of the remainders is 1 + 0 + 0 + ... + 0 = 1.

So, the remainder when 1! + 2! + 3! + ... + 100! is divided by 2 is 1.

Let's analyze the factorials modulo 2:

1! = 1 ≡ 1 (mod 2)

2! = 2 ≡ 0 (mod 2)

3! = 6 ≡ 0 (mod 2)

4! = 24 ≡ 0 (mod 2)

Notice that for any n ≥ 2, n! will contain a factor of 2, meaning that n! will be an even number. Therefore, n! ≡ 0 (mod 2) for all n ≥ 2.

So, the sum becomes:

1! + 2! + 3! + ... + 100! ≡ 1 + 0 + 0 + ... + 0 (mod 2)

1! + 2! + 3! + ... + 100! ≡ 1 (mod 2)

Therefore, the remainder when 1!+2!+3!+⋯+100! is divided by 2 is 1. Final Answer: The final answer is 1