Question
For an integer n, n!=n.(n-1).(n-2). ... 3.2 .1. Then
1! +2!+3!+...+ 100! when divided by 5 leaves remainder _________.Solution
ATQ, Factorials Modulo 5: 1! = 1 → Remainder = 1 2! = 2 → Remainder = 2 3! = 6 → Remainder = 1 (since 6 ≡ 1 mod 5) 4! = 24 → Remainder = 4 (since 24 ≡ 4 mod 5) 5! and beyond → Divisible by 5, so remainder = 0 Sum of Factorials Mod 5 = 1! + 2! + 3! + 4! = 1 + 2 + 1 + 4 = 8 8 mod 5 = 3 Hence, the remainder is 3.
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