For which of the following n is the number 274 + 22058 + 22n a perfect square?
option
(a) 2012
(b) 2100
(c) 2011
(d) 2020

• 2^74 +2^2058+2^2n = K^2
  2^74 +2^2058+2^2n = (2^37)^2+2^2058+(2^n)^2
  We try to write this expression as (a+b)^2=a^2+2ab+b^2
  Now a = 2^37, 2ab = 2^2058 and b = 2^n
  Substituting the value of a in 2ab, we get b = 2020
  
• 11 years agoHelpfull: Yes(10) No(13)
• This can be simplified as
  (2^37)^2+ 2*(2^37)*2^n + (2^2n)
  So 38+n=2058
  => n=2020.
• 11 years agoHelpfull: Yes(5) No(0)
• d) 2020
  2^74 +2^2058+2^2n = K2
  2^74 +2^2058+2^2n = (237)2+22058+(2n)2
  We try to write this expression as (a+b)2=a2+2ab+b2
  Now a = 237, 2ab = 22058 and b = 2n
  Substituting the value of a in 2ab, we get b = 2020
  
• 11 years agoHelpfull: Yes(1) No(5)
• hw u got 237?
• 11 years agoHelpfull: Yes(0) No(2)
• 2^74+2^2058+2^2n
  =(2^37)^2+2*(2^37)*(2^2020)+(2^2020)^2 /*since (37+1=38 & 2058-38=2020)*/
  =(2^37+2^2020)^2
  ans=2020
• 11 years agoHelpfull: Yes(0) No(0)
• a2+b2+2ab
  a=2^37
  b=2^n
  substitute in 2ab
  2^1*2^37*2^n=2058
  38+n=2058
  n=2058-38
  ans=2020
• 11 years agoHelpfull: Yes(0) No(0)