(2^1096+2^2248+2^2n)find the value of n so that the value will be perfect square(a)2011(b)2012(c)2010(d)2008

• None of given options is correct.
  (2^1096+2^2248+2^2n) = ((2^548)^2 +2^(1+n+548)+2^2n)
  
  so
  1+n+548 = 2248
  n= 2248-549 = 1699
• 11 years agoHelpfull: Yes(6) No(8)
• substituting the values of n=2011,2012,2010,2008 in the question... we get the total solutions as 7366 7368 7364 7360...now 7360=(7+3)(6+0)=60 not a perfect square whereas 7364=(7+3)(6+4)=100 a perfect square so the answer is (c)2010
• 11 years agoHelpfull: Yes(1) No(7)
• (2^1096 + 2^2248 + 2^2n)
  there is + between the numbers not * how you giving your answer
• 11 years agoHelpfull: Yes(0) No(4)
• 7364 is not a perfect squre
• 11 years agoHelpfull: Yes(0) No(5)
• sorry friends for the above answer...but this answer is mostly right..
  
  put the values of n in the problem from the options...then the series is
  1096 , 1096 + 56 ,1096 + 13*56 =4016 whn n=2008 then 2n=4016....so answer is (d)2008
• 11 years agoHelpfull: Yes(0) No(4)