for which of the following n is the number (2^74)+(2^2058)+(2^2n) a perfect square..
a. 2010
b. 2011
c. 2012
d. 2100

• 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.
• 12 years agoHelpfull: Yes(20) No(1)
• none of given options
  
  (2^74)+(2^2058)+(2^2n)
  
  here for perfect square
  2^2058 = 2*(2^37)* 2^n= 2*(2^37)* 2^2020
  
  so n=2020
• 12 years agoHelpfull: Yes(8) No(2)
• it is a dummy question.
  no answer matches.
  to solve it we know that a perfect square should have 1, 6 ,4, 9 at the last digit. so find last digit of first two terms like this 2^5 gives 2 at units place so uptill 2^70 it will be 2 and then find for rest 4 it gives 6 so unit place digit for 1st term is 6*2=12...take only first digit..solve second term like this and then by putting options in the third term you will find that all options are wrong.
• 12 years agoHelpfull: Yes(3) No(6)
• 2020 is the right ans
  
• 12 years agoHelpfull: Yes(3) No(3)