Number Theory

 

Find the remainder when $1! + 2! + 3! + \dots + 100!$ is divided by $2$.    

 

Every factorial will have a "times 2" in it.  Every factorial that is, except 1!.    

 

All the factorials added together, except 1!, will sum to an even number.    

 

Therefore, the remainder when the total sum is divided by 2 will be that 1.    

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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

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.