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Q61
(CAPF/2024)
Miscellaneous & General Knowledge › Important Days, Places & Events › Important Days, Places & Events
The remainder, when 1 + (1 × 2) + (1 × 2 × 3) + ... + (1 × 2 × 3 × ... × 500) is divided by 8, is
Result
Your answer:
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·
Correct:
A
Explanation
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.
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