1!+2!+3!...+50! Divided by 5! Remainder will be ?

• 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.
• 12 years agoHelpfull: Yes(22) No(4)
• 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)
  =33/120=33
• 12 years agoHelpfull: Yes(9) No(1)
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• 12 years agoHelpfull: Yes(4) No(1)
• 1!+2!+3!+4!=33 after this we get a zero in all the factorials. so the remainder is (no end with 0+ 33)/5 = 3
• 12 years agoHelpfull: Yes(3) No(2)
• only !1,!2,!3,!4 will not be divisible by 5 rest have 5 with them all will divisible by 5
  so !1+!2+!3+!4=33
  now 33%5=3
  ans will 3
• 12 years agoHelpfull: Yes(0) No(2)