(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

• to be a perfect square we can rite given term into foll. format (a+b)^2=a^2+2*a*b+b^2
  (2^548)^2 + 2*(2^548)*(2^n)+(2^n)^2
  comparing it ith given term,,,, 2248=1+548+n
  n=1699 is correct
• 12 years agoHelpfull: Yes(8) No(0)
• (2^1096 + 2^2248 + 2^2n)
  kanak saxena ji there is + between numbers not * how you give your answer
• 12 years agoHelpfull: Yes(3) No(2)
• ans is (a) by hit & trial method.
  to be a perfect sq. a nos power should be multiple of 2, obtion b,c,d satisfy this statement.
  2^(1096 + 2248 + 2n) = 2^(3344+ 2n)....(1)
  n= 2011, 2012 , 2010 , 2008.
  by putting d value of n in eqt 1, we get 7366 , 7368 ,7364 and 7360 resp..
  by dividing this sum by d cyclicity of 2 i.e 4 we get integer 4m obtion b, c,and d.
  oddmann out, obtion 'a' is differ 4m other , so answer is a. ;-)
• 12 years agoHelpfull: Yes(1) No(12)