1!+2!+3!...+50! when divided by 5!, the remainder is?

• 5!=120 now look at the sum it is clear that
  after 5! all are divisible by 5!(120) so
  take the sum of first four terms i.e.,
  1!+2!+3!+4!=33
  therefore rem=33.
• 10 years agoHelpfull: Yes(55) No(1)
• 33 will be the remainder.
• 10 years agoHelpfull: Yes(7) No(1)
• 33
  no need of finding remainder after 5!
  (1!+2!+3!+4!)%5!=33%120=33
• 10 years agoHelpfull: Yes(4) No(1)
• remainder is 33
• 10 years agoHelpfull: Yes(4) No(0)
• 5!/5! + 6!/5! + .... gives no remainders, so you only have to worry about the first 4 terms.
  
  (1! + 2! + 3! +... 50!)/5!
  = 1!/5! + 2!/5! + 3!/5! + 4!/5! + ... 50!/5!
  = (1! + 2! + 3! + 4!)/5! + .... 50!/5!
  
  (1! + 2! + 3! + 4!)/5!
  = (1 + 2*1 + 3*2*1 + 4*3*2*1)/5!
  = (1+2+6+24)/5!
  = (33)/5!
  = (33)/(5*4*3*2*1)
  = (11)/(5*4*2*1)
  = 11/(40)
  
  Remainder 11
• 10 years agoHelpfull: Yes(3) No(18)
• 33 how..?can anyone explain?
• 10 years agoHelpfull: Yes(1) No(5)
• rem is 33%5=3
• 10 years agoHelpfull: Yes(1) No(4)
• 1!+2!+3!+4!=33
  rem=33.
• 10 years agoHelpfull: Yes(0) No(0)