x lies from 121 < X >1331, x^2+1 when divided by 11 what will not be the remainder .
A.1 b.4 c.6 d.8

• answer is d.
  i have such used hit and trial method.
  part 1:
  see,x is greater than 121 means (11)^2 and less than 1331 i.e less than than (37)^2 [as 37^ is 1369,so take value till (36)^2=1296]
  =>x could be from 12^2 to 36^2
  
  part 2:
  (x^2+1)/11
  (i)put x=12 in the equation
  ( 12^2+1)/11-->R=2(so not beneficial as it not in the option)
  go on checking....X=13,14,15,16....
  (ii)x=15
  R=6
  (iii)x=16
  R=4
  (iv)think when would be remainder 0 ,it has to be 11 ka square or multiple of 11 ka square,as 11 ka square cannot be taken,next multiple of 11 would be 22,put x=(22^2+1)/11-->R=1
  (u will have to think a bit)
  now the only left option is d i.e 8
  -----------------------------------------------
  if u know theoretically not by hit and trial plz share ur knowledge
• 7 years agoHelpfull: Yes(8) No(2)
• x lies between 11^2 and 11^3
  SUPPOSE
  x=11^2
  then x^2+1 become 1332
  1332/11 gives rem 1
  BUT X CANT BE 11^2
  SO WRONG REM IS 1
• 7 years agoHelpfull: Yes(1) No(5)
• i have made a program for this question and then only 8 is not comming in the reminder
  so the answer is d)8
• 7 years agoHelpfull: Yes(1) No(0)
• For any range x^2+1 when divided by 11 forms a cyclic pattern which is
  1 2 5 10 6 6 10 5 2 1 1 2 5 10 6 6 10 5 2 1 and so on.
  So we can see that 8 never occurs
• 7 years agoHelpfull: Yes(1) No(3)
• 11*11
• 7 years agoHelpfull: Yes(0) No(2)
• #inlcude
  int main()
  {
  int x,r;
  x=122;
  while(x
• 7 years agoHelpfull: Yes(0) No(1)