The remainder obtained when \sum^{100}_{(n-1)}(n!) is divided by 6 is

AP ICET 2015 Paper

Question : 127 of 200

Marks: +1, -0

The remainder obtained when

∑↖{100}↙{(n-1)}(n!)

is divided by 6 is

Solution:

∑↖{100}↙{n=1}(n!)=1!+2!+3!+4!+....+100!

3!=3×2×1,4!=4×3!,5!=5×4×3!

and so on.
So,

(3!+4!×5!+..........+100!)

is divisible by 6
∴ Required remainder

=1!+2!=1+2=3

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