Let T be a set of integer{2,4,8,32......2^n} and S be a subset of T,such that the sum of no two element of S is greater than 2^n-1. Let M be the maximum number of elements S can have for a given n. What will be the value of M?
A. n-1
B. n-2
C. n/2
D. n

• If it is 2^(n-1),
  n-2 should be the answer..
  let n=5 so T={2,4,8,32,64}
  Now the condition is that, sum of NO two elements of the subset S should be greater than 2^(n-1) the subset obviously shuold be S={2,4,8} which has n-2 elements
• 11 years agoHelpfull: Yes(46) No(1)
• n-1....as after excluding the last element the condetion will be true for any two elements in the set
• 11 years agoHelpfull: Yes(9) No(11)
• brain is right its n-2
• 11 years agoHelpfull: Yes(4) No(0)
• let n=6
  T={2,4,8,32,64}
  S={2,4,8} as S={sum must be less than 2^n-1}
  therfore, M=n-2
• 11 years agoHelpfull: Yes(4) No(0)
• let n=2
  t={2,4}
  n-1=2-1=1
  2^2-1=2
  M={2}
  max elements =n-1
  
• 11 years agoHelpfull: Yes(2) No(12)
• take n=5.
  then 2^n-1 = 16.
  thereore s={2,4,8}(since in s now no two sum is greater than 16)
  now,m=3(since s contain 3 elements only),
  but,we took n=5 at start,so just verify options
  we get B.N-2 is correct,as n-2 = 5-2 = 3 = m.
• 11 years agoHelpfull: Yes(1) No(0)
• n-2 is the ans
• 10 years agoHelpfull: Yes(1) No(1)
• plz elaborate ur answer
• 11 years agoHelpfull: Yes(0) No(0)
• x+x =2x.
  (x+anything
• 11 years agoHelpfull: Yes(0) No(0)
• x+x =2x.
  (x+anything
• 11 years agoHelpfull: Yes(0) No(0)
• I m having elitmus exam on this sunday. Cn you please send all the material related to this exam nd specially cryptarithmetic problems wid procedure. My email id is rohitgoyal.8214@gmail.com
  
• 11 years agoHelpfull: Yes(0) No(0)