Given 2^74 + 2^2074 + 2^2n. For what value of(among the options)n will the sum be a perfect square?

• i think n=2036
• 12 years agoHelpfull: Yes(12) No(5)
• n=1075/2 = 537.5
  
  then
  2^74 + 2^2074 + 2^2n = (2^37)^2 +(2^1037)^2 +2* (2^37)*(2^1037)
  = (2^74 +2^1037)^2
  There may be other possibilities also.
  but here options are not given. pls check.
  
• 12 years agoHelpfull: Yes(12) No(0)
• guys..its a dummy questn..dont mark any optn..
• 12 years agoHelpfull: Yes(7) No(1)
• we know, (a+b)^2= a*a+b*b+2a*b
  so 2^74+2^2074+2^2n= (2^37)^2 + 2*2^37*2^n + (2^n)^2
  So, we get 38+n=2074
  therefoe n = 2036.
• 12 years agoHelpfull: Yes(6) No(0)
• ans should be 2036
  use (a+b)^2
  
  a=2^37 b= 2^n 2*a*b =2*2^37*2^n ie 2^1+37+n =2^2074
  
  38+n=2074
  n=2036
• 12 years agoHelpfull: Yes(2) No(0)
• i think
  974..
  
• 12 years agoHelpfull: Yes(0) No(3)
• we only takes unit place power/4 and remainder place it unit digitso
  2^2+2^2+2^2*3/2=16
• 12 years agoHelpfull: Yes(0) No(3)
• 2^2n=2^2*2036
  
  so,n=2036
• 12 years agoHelpfull: Yes(0) No(0)