What is the trick for solving following remainder questions?
1. 123123123123/11 = rem
2. 989898/101= rem?
3.(11111*22222*33333)/5 = rem?

• (1). When 123123 is divided by 11 we get 11193 as quitent and 0 as remainder.
  
  It means that 123123 is completely divisible by 11 , if next six digits are
  
  also same then that term will also be divisible by 11 completely.
  
  i.e. when we divide 123123123123 by 11 we get 0 as a remainder.
  
  (2). 9898 is completely divisible by 101 , it means remainder will be 98
  
  (3). (11111*22222*33333)/5
  using remainder theorem,
  when we diide 11 by 5 we have 1 as remainder.
  Similiarly when we divide 11111 by 5 we get 1 as remainde.
  
  Now, dividing 22 by 5 gives 2 as remainder,
  similiarly 22222 will give 2 as remainder.
  
  And 33333 will give 3 as remainder.
  
  now, 1*2*3/5 = 6/5 = 1
  
  Therefore 1 is the remainder
• 9 years agoHelpfull: Yes(0) No(0)
• hello sudhir, thnx for explaining
  i think in last 11111*22222*33333 they are already divided by 5 seprately and remainder comes out to be 1*2*3 = 6
  6 should be remainder ??
  or i am mistaking somewhere?
• 9 years agoHelpfull: Yes(0) No(0)