1 2 3 50 Divided by 5 Remainder will be 0 or 33

from 5! the sequence will have 0 in the units place,for which the remainder turn out to be 0.
the rest will be 1!+2!+3!+4!
11 years agoHelpfull: Yes(12) No(2)
Clearly 5! onwards each is divisible by 5!.
SO the remainder will come from the division (1!+2!+3!+4!) divided by 5!
Now (1!+2!+3!+4!) = 1+2+6+24 = 33 and 5! = 120 so the remainder is 33.
11 years agoHelpfull: Yes(11) No(0)
after 5! the sum upto 50! is divisible by 5!=120 ,
but 1!+2!+3!+4! =33 is not divisible by 5!

so remainder is 33
11 years agoHelpfull: Yes(3) No(0)
1!+2!+3!+4!=33(rest numbers will be cancelles as it contains equals to or greater than 5!)
hence the remainder is 33
11 years agoHelpfull: Yes(2) No(0)
1+2+6+24=33
ans
11 years agoHelpfull: Yes(1) No(0)
remainder will be 0
as we know sum of 1to 9 integers is equal to 55 which is divisible by 5
11 years agoHelpfull: Yes(0) No(8)
1!+2!+3!+4!=33
33%5!=33
11 years agoHelpfull: Yes(0) No(0)

1!+2!+3!...+50! Divided by 5! Remainder will be ? (0 or 33)