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

• since x should be greater than 121, let us suppose that 'y' is no. which is greater than 0,
  now we can write, x^2+1 as (121+y)^2 +1.
  (y^2 + 121^2 + 2*121*y + 1)/11
  y^2/11 + 121^2/11 + 2*121*y/11 + 1/11
  y^2/11 + 0 + 0 + 1 ( remainder)
  y^2/11 + 1
  y^2 can be 1,4,9,16,25......
  after dividing by 11,remainder we got 1,4,6 but not 8.
  so ans is d.
• 7 years agoHelpfull: Yes(19) No(6)
• Vikash Please clarify
  
  since x should be greater than 121, let us suppose that 'y' is no. which is greater than 0,
  now we can write, x^2+1 as (121+y)^2 +1.
  (y^2 + 121^2 + 2*121*y + 1)/11
  y^2/11 + 121^2/11 + 2*121*y/11 + 1/11
  y^2/11 + 0 + 0 + 1 ( remainder)
  y^2/11 + 1
  y^2 can be 1,4,9,16,25......
  after dividing by 11,remainder we got 1,4,6(is it 6 or 9),5.... and we should add +1 also to this right because of 1/11 ??
• 7 years agoHelpfull: Yes(5) No(1)
• option d>8
• 7 years agoHelpfull: Yes(1) No(0)
• how u got ths
• 7 years agoHelpfull: Yes(0) No(0)