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

LiIIiam0216 Jul 27, 2024

#1**+1 **

Let's make some observations.

Starting with \(2!\), every other term after that is even, since it contains 2.

Since we know that \(even + even = even\), then we have \(2! + 3! + \dots + 100! = even\)

Thus, the sequence \(2! + 3! + \dots + 100! \equiv 0 (\mod 2)\)

However, we forgot about 1 term. Since we have 1! = 1, and 1 is odd, and we know \(odd+even=odd\), then we have

\(1! + 2! + 3! + \dots + 100! \equiv 1(\mod 2)\)

Thus, the answer is 1.

Thanks! :)

NotThatSmart Jul 29, 2024