(2*6!+21*6!)/(14*7!+14*13!) find remainder?

• 2*6!+21*6! can be written as 23*6!.
  denominator can be written as 14*7!*(1+8*9*10...*13)= 14*7*6!*(1+8*9*10...*13)
  if we divide we can cancel out 6!.
  now 14*7(1+8*9*10*..*13) is a large not divisible by 23.. so 23 is remainder.
• 11 years agoHelpfull: Yes(38) No(8)
• remainder would be 23*6!=16560
• 11 years agoHelpfull: Yes(13) No(7)
• as denominator is greater than numerator ans is 23*6!
  
• 11 years agoHelpfull: Yes(6) No(1)
• remainder must be 23
• 11 years agoHelpfull: Yes(3) No(3)
• ans is - 23
• 11 years agoHelpfull: Yes(2) No(10)
• mehar can u pls elavorate...
• 11 years agoHelpfull: Yes(1) No(1)
• ans - 23 take out common factorial
• 11 years agoHelpfull: Yes(1) No(2)
• remainder=7
• 10 years agoHelpfull: Yes(0) No(0)
• as numerator 23*6! is smaller you should not cancel 6! as it will divide remainder by 6!. So 23*6! is answer.
  
• 9 years agoHelpfull: Yes(0) No(0)