To determine the remainder of \(1! + 2! + 3! + \dots + 100!\) when divided by \(2\), we need to consider the parity of each factorial term (\(n!\)).
A number \( n! \) (the factorial of \( n \)) is the product of all positive integers from \(1\) to \(n\). For \( n \geq 2 \), \( n! \) includes at least one factor of \(2\) because it contains both \(1\) and \(2\) as factors, hence \( n! \) for \( n \geq 2 \) is always even.
We now list the values of the factorials up to \(2!\):
\[
\begin{aligned}
1! &= 1, \quad &(\text{odd})\\
2! &= 2, \quad &(\text{even})\\
n! &= \text{even}, &\text{for } n \geq 2
\end{aligned}
\]
Hence, all factorial terms \(2!, 3!, 4!, \ldots, 100!\) are even and contribute \(0\) to the sum when considering their remainder upon division by \(2\).
The sum \(1! + 2! + 3! + \dots + 100!\) is simplified by focusing on the parity sums:
\[
1! + 2! + 3! + \cdots + 100! \equiv 1 \pmod{2}
\]
This is because all terms from \(2!\) to \(100!\) are even (contributing \(0\) in modulo \(2\)), and the only odd term is \(1!\).
Thus, the remainder when \(1! + 2! + 3! + \cdots + 100!\) is divided by \(2\) is
\[
\boxed{1}
\]
Find the remainder when $1! + 2! + 3! + \dots + 100!$ is divided by $2$.
1 answer