What is the remainder when \(1! + 2! + 3! + ... + 100!\) is divided by 8?
Answer: A
We need to find the remainder of the sum when divided by 8. Let's look at the remainders of the first few factorials. \(1! = 1\) (rem 1). \(2! = 2\) (rem 2). \(3! = 6\) (rem 6). \(4! = 24\). Since 24 is a multiple of 8, the remainder is 0. For any \(n \ge 4\), \(n!\) contains the factors 4 and 2, so it is a multiple of 8. The remainder for all subsequent factorials will be 0. We only need to sum the remainders of the first three factorials: \(1 + 2 + 6 = 9\). The remainder of \(9 \div 8\) is 1.