ParthKohli
  • ParthKohli
How will we solve this? What will be the remainder after we divide 50! + 49! + 48!.......2! + 1! by 5! ?
Mathematics
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ParthKohli
  • ParthKohli
How will we solve this? What will be the remainder after we divide 50! + 49! + 48!.......2! + 1! by 5! ?
Mathematics
chestercat
  • chestercat
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ParthKohli
  • ParthKohli
@FoolForMath I need your genius here
ash2326
  • ash2326
50!.... 5! each of these have a 5! so remainder will be 4!+3!+2!+1!=24+6+2+1=33
ParthKohli
  • ParthKohli
I am not sure how you got it :|

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anonymous
  • anonymous
It's easy, Starting from 5! every factorial is a divisible by 5.
anonymous
  • anonymous
So, just ignore the rest.
ParthKohli
  • ParthKohli
Oh yes!
ash2326
  • ash2326
Did you understand @ParthKohli ?
ParthKohli
  • ParthKohli
Yes! that's the principle of how we divide \(\Large {6! \over 5!}\) and we get \(6\)... so 6 is the quotient and there is no remainder! ^.^
anonymous
  • anonymous
|dw:1340545398596:dw|
anonymous
  • anonymous
how about equation 1!+2!+3!+...+n!=n^2 how many answers does it have n is nature?
ParthKohli
  • ParthKohli
Thank you all!

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