2^74+ 2^2058+ 2^2n for which of the following n is a perfect square?
2012 , 2011 , 2010 ,2100

• Ans: It is a dummy question.
  soln:
  we know (a+b)^2= (a^2)+(b^2)+2ab
  form the question,
  a^2=(2^74), a=(2^37), 2ab=2^2058, b^2=(2^2n)
  therefore,2ab=2*(2^37)*(2^x)
  find the value of x. 2ab=(2^((1+37+x)=2058)
  solving this we get, x=2020.
  2^(b^2)=(2^(2020)^2).
  2^2n=(2^2(2020))
  so n=2020.
  no option is given. so it is a dummy question.
• 11 years agoHelpfull: Yes(28) No(2)
• (2^37)^2=2^74
  2*(2^37)*2^2020=2^2058
  (2^37)^2+(2*(2^37)*2^2020)+(2^2020)^2 which is expressed in (a+b)^2 form where a=2^37
  b=2^2020
  therefore (2^2020)^2=2^2n
  n=2020
• 11 years agoHelpfull: Yes(6) No(2)
• answer is 2011
• 11 years agoHelpfull: Yes(3) No(7)
• if it is a dummy question, so what option we should choose in test??
• 11 years agoHelpfull: Yes(2) No(0)
• answer is 2010...
• 11 years agoHelpfull: Yes(0) No(5)