Remainders in Factorial Problems :

1.What is the remainder when 1! +2! +3! +4! ...+100! is divided by 24?

24=6*4 or 3*2^3
we know 4! =24 & all the factorial values after 4! Will be divisible by 24  (for example 5! =5*4*3*2*1=120 divisible by 24 , 6! =6*5*4*3*2*1=720 it is also divisible by 24 so further all the factorial values will be divisible by 24 ) . On dividing all these factorials by 24 we would obtain 0 as the remainder.

But we get a remainder in 1! +2! +3! Because these 3 values are not divisible by 24
So remainder for the problem will be
1! +2! +3! =1+2+6=9

2.What is the remainder when 1!+2!+3!+4!+5!+.......+50! is divided by 5!
   
 Like previous problem from 5! Upto 50! All the values are divisible by 5! And will give 0 as remainder.
Remainder for left out values will be 1! +2! 3! +4! =1+2+6+24=33.

3.What is the remainder when 1! + 2! + 3! … 100! is divided by 18? :

18=6*3 or 3^2*2  so we have to get two 3's and one 2 to get a number divisible by 18 .

From 6! Upto 100! All the values are divisible by 18 . Except the remainder achieved through
1! +2! 3! +4! +5! =1+2+6+24+120
                          Remainder =153