find remainder
(1!+2!+3!+........+99!)/720
please explain your method clearly

• (1!+2!+3!+........+99!)/720
  
  (1!+2!+3!+........+99!)/6!
  
  from 6! onwards each term is divisible by 720 = 6!
  hence ( 1!+2!+3!+4!+5! ) / 6! gives reqd remainder
  => 1+2+6+24+120 / 6!
  => 153 / 6!
  rem = 153
• 10 years agoHelpfull: Yes(7) No(0)
• 1!=1
  2!=2
  3!=6
  4!=24
  5!=120
  6!=720
  7!=720*7
  
  after these all factorials will be multiple of 720 hence remainder lies beacause of
  1!+2!+3!+4!+5!=1+2+6+24+120
  =153
• 10 years agoHelpfull: Yes(3) No(0)