14) The remainder when 1!+2!+3!...+50! divided by 5! will be

• here from 5! to 50! divided by 5!
  
  so we have to find just 1!+2!+3!+4!=33 so reminder=33
• 9 years agoHelpfull: Yes(16) No(0)
• 6!/5!=(rem.)=0
  7!/5!=0
  .
  .
  .
  50!/5!=0
  
  1+2+6+24/120=33/120
  rem.=33
• 9 years agoHelpfull: Yes(4) No(0)
• factoria above than 4! will be fully divisible
  
  1+2+3*2+4*3*2=33
  33/5 remainder 3
• 9 years agoHelpfull: Yes(1) No(0)
• remainder=11
  
  (1!+ 2!+ 3!+ .... +50!)/5!
  
  =((1! + 2! + 3! + 4!)/5!) + ((5! + 6! + ... + 50!)/5!)
  
  =(1+2+6+24)/5! {since the latter part is divisible by 5!, hence generates no remainder}
  
  =33/120
  =11/40.
  therefore, remainder=11.
• 9 years agoHelpfull: Yes(1) No(3)