4(4!)+5(5!)+6(6!)+7(7!)+.......+19(19!)

• correct Q: find remainder of (4(4!)+5(5!)+6(6!)+......+19(19!)) /64.
  sol:
  5-1(4!)+6-1(5!)+7-1(6!)+.....+20-1(19!)/64
  5.4!-4!+6.5!-5!+7.6!-6!+....20.19!-19!/64
  5!-4!+6!-5!+7!-6!+...........+20!-19!/64
  (-4!+20!)/64
  
  20! is divisible by 64, therefore remainder is 64-4!=40
  
• 9 years agoHelpfull: Yes(12) No(0)
• 8*8!+9*9!....... will be completely divisible by 64 so we have to concern about first 4 terms so
  4(4!)+5(5!)+6(6!)+7(7!) is 40296 when divided by 64 will gives 40 so answer is 40
• 9 years agoHelpfull: Yes(6) No(0)
• 20(20!) - 1 -4 -18
• 9 years agoHelpfull: Yes(0) No(1)
• n*n! ka summation (n+1)! - 1
  20!-1 -(4!-1) = 20!-4! leave the first ans submitted by me
• 9 years agoHelpfull: Yes(0) No(1)
• option kya tha???????
• 9 years agoHelpfull: Yes(0) No(0)
• here we have to find remainder if it will divide by 64.
  
• 9 years agoHelpfull: Yes(0) No(1)
• =(4.4! + 5.5! + 6.6! +7.7!)/64
  take 4! common
  4!(4 + 5.5 + 6.6.5 + 7.7.6.5)/64
  =4!(4+25+180+1470)/64
  =4!*1679/64
  =4*3*2*1679/64
  =3*1679/8
  =5037/8
  5032 is divisible by 8 so 5 would be the remainder .
• 9 years agoHelpfull: Yes(0) No(4)