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 _________.

Q: Answer-For an integer n, n!=n.(n-1).(n-2). ... 3.2 .1. Then 1! +2!+3!+...+ 100! when divided by 5 lea...

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.

B 1

C 2

D 3

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For an integer n, n!=n.(n-1).(n-2). ... 3.2 .1. Then 1! +2!+3!+...+ 100! when divided by 5 leaves remainder _________.

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