What is the remainder when \(7! + 8! + 9! + ... + 100!\) is divided by 6?
Answer: A
We need to check the remainder of each factorial when divided by 6. \(n! = 1 \times 2 \times 3 \times ... \times n\). For any \(n \ge 3\), the factorial \(n!\) contains the factors 2 and 3. Therefore, for \(n \ge 3\), \(n!\) is a multiple of \(2 \times 3 = 6\). This means the remainder of \(n! \div 6\) is 0 for all \(n \ge 3\). The given series starts from 7!. Since 7! and all subsequent factorials are multiples of 6, their remainder when divided by 6 is 0. The sum of zeroes is 0. So the final remainder is 0.