For which of the following “n” is the number 2^74 + 2^2058 + 2^2n a perfect square?

a. 2012
b. 2100
c. 2011
d. 2020

• 2^74+2^2058+2^2n= (2^37)^2+ (2^n)^2+ 2^2058
  nw compare it wid (a+b)^2=a^2+b^2+2*a*b
  a=2^37 and b=2^n
  thus 2*(2^37)*(2^n)=2^(38+n)=2^2058
  38+n=2058;
  n=2020
• 11 years agoHelpfull: Yes(45) No(0)
• ans will be option (c)
  for the perfect square we have to check 74%4 then remainder will be 2
  2058%4 remainder will be 2
  we should put n= 2011,then 2n=4022,again dividing 4022%4,remainder will be 2
  then for the perfect square we should take 2^2 common,which is to a perfect square
• 11 years agoHelpfull: Yes(7) No(6)
• make the exp in the form of a^2+b^2+2ab here a=2^37 and b=2^n then 2ab=2^2058 putting the the value of a and b. 2^n=(2^2058)/2^38 by comparing the power of n we get n=2020 ANS
• 11 years agoHelpfull: Yes(4) No(0)
• ans is 2020
• 11 years agoHelpfull: Yes(0) No(0)