Number Theory

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 $1!=1$, 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! :)