The remainder, when 1 + (1 × 2) + (1 × 2 × 3) + ... + (1 × 2 × 3 × ... × 500) is divided by 8, is

The given series is the sum of factorials: 1! + 2! + 3! + ... + 500!

We need to find the remainder when this sum is divided by 8. Let's calculate the values of the first few factorials:

• 1! = 1
• 2! = 1 × 2 = 2
• 3! = 1 × 2 × 3 = 6
• 4! = 1 × 2 × 3 × 4 = 24

Notice that 4! (which is 24) is perfectly divisible by 8. Because every factorial from 4! onwards (5!, 6!, etc.) contains 4! as a factor, they are all completely divisible by 8 and leave a remainder of 0.

Therefore, we only need to consider the sum of the terms before 4!:

1! + 2! + 3! = 1 + 2 + 6 = 9

When 9 is divided by 8, the remainder is 1.