(2^1096 + 2^2248 + 2^2n) find the value of n so that the value will be perfect square (a)2011 (b)2012 (c)2010 (d)2008

• find the unit digit of 2^1096
  wch cms out 2 b = 6
  nw again unit digit of 2^2248= 6
  add these unit digits= 6+6=12
  now we knw any no. wch ends wd 2 is not a perfect square.
  
  consider last term: 2^2n= 4^n
  if n=2011 then unit digit of 4^2011= 4
  add 12+4 = unit digit= 6
  it can make a perfect square
  
  if n=2012, then unit digit of4^2012 = 6
  adding 12+6+unit digit= 8
  no.ending wd 8 can not b a perfect square.
  ||ly wd the other options. :)
  so ans is 2011
• 11 years agoHelpfull: Yes(18) No(7)
• ansr:2011
  just calculate unit digit in b,c,d it will come 3 nd the number which end with 3 cant be perfect square
• 11 years agoHelpfull: Yes(13) No(36)
• to find the value of n, just look at options first
  2011, 2012,2010 are not divisible by 8 but 2008 is divisible by 8......
  also 1096 and 2248 is also divisible by 8
  hence ans is (Id) 2008
• 11 years agoHelpfull: Yes(3) No(13)
• ans is 2012
  as 2^1096 give unit digit 6
  2^2248 give umit digit 6
  and now we calculate unit digit of third i.e 2^2*(2011) give unit digt 2
  so (6*6*2= 72 here unit digit of 2 is not a perfect sq)
  so take now 2^2*2012 which give unit digit 4 so now (6*6*4=144)
  hence 144 unit digit is 4 so perfect sq..
• 11 years agoHelpfull: Yes(1) No(9)