what is the remainder when 1!+2!+3!+...+199! is divided by 21?

• each term from 7! to 199! is completely divisible by 21
  so (1!+2!+...+6!)/21 gives the required remainder
  =(1+2+6+24+120+720)/21=873/21 which gives remainder 12
  ans is 12
• 10 years agoHelpfull: Yes(46) No(0)
• 12 Ans.
  because 7! are divided by 21.so 8!to 199!are also divided by 21.
  only we check 1!+2!....6!=873/21 gives remainder 12.
  
• 10 years agoHelpfull: Yes(7) No(0)
• (1!+2!+3!+4!+5!+6!)/21= rem..12
  rest of terms are divisble by 21
• 10 years agoHelpfull: Yes(3) No(1)
• bhaag milkha bhaag
  
• 10 years agoHelpfull: Yes(2) No(1)
• each term from 7! to 199! is completely divisible by 21
  so (1!+2!+...+6!)/21 gives the required remainder
  =(1+2+6+24+120+720)/6=873/6 which gives remainder 3
  ans is 3
• 10 years agoHelpfull: Yes(1) No(4)
• ans is 12
  
• 10 years agoHelpfull: Yes(1) No(0)
• 12
  it is divisible 4m 7! onwards
  so remainder of 1!+2!+3!+4!+5!+6! is the answer
  
• 10 years agoHelpfull: Yes(0) No(0)
• from 6! onwards all the terms will be divisible by 21.
  as we know 21=7*3.
  7!,8!,......199! all have 7and3 so , it will be completely divisible.
  
  1!+2!......6!= 873%21 = 12.
  so the remainder will be 12.
• 10 years agoHelpfull: Yes(0) No(0)