+2669 Find the remainder when $1! + 2! + 3! + \dots + 100!$ is divided by $2$.
+1897 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! :)
