2^1096+2^2248+2^2n
for what value of n it become perfect square
a)2008
b)2010
c)2011
d)2012

• answer will be 2011.
  see unit digit of 2^1096 is 6((2^4)^274)
  and unit digit of 2^2248 is 6 also...
  so sum of those digit is 12..which is not a prfct sqr..
  now if unit digit of 2^2n becomes 4..den 12+4=16 can be a prfct sqr..
  if we take the option 2008.2^2n will be 2^4016..unit digit of this is 2..
  which is not correct..if we take 2011..2^2n will be 4^2011..unit digit of ds is 4.
  now the sum is 12+4=16..which is a prfct sqr..
  
• 11 years agoHelpfull: Yes(9) No(2)
• wrong question
  (a+b)^2=a^2 + b^2 + 2ab
  if its around 2000 then
  it cant be 2ab(2^(548+1124+1))
  also 2^1096 cant be 2ab(
• 11 years agoHelpfull: Yes(4) No(2)
• 1096
• 11 years agoHelpfull: Yes(0) No(2)
• 1096=8*137
  2248=8*281
  so,
  2008=8*251....
  
  therefore,
  (A) is the ans
• 11 years agoHelpfull: Yes(0) No(4)