1! + 2! + 3! + 4! + 5! + 6! + …. 100! when divided by 5, would leave remainder
Explanation:
All the terms in the series from 5! onwards are divisible by 5.
∴ Remainder will be the same as the remainder when (1! + 2! + 3! + 4!) is divided by 5
⇒ 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33.
∴ The remainder will be R(33/5) = 3.
Hence, option (d).
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