Every day in his business, a merchant had to weigh amounts from one pound to one hundred and twenty-one pounds, to the nearest pound. To do this, what is the minimum number of weights he needs and how heavy should each weight be?

• The minimum number of weights required is five and these should weigh 1, 3, 9, 27 and 81 pounds.
  
  
• 12 years agoHelpfull: Yes(4) No(1)
• The solution wud be in powers of 3 ..ie 3^0 , 3^! , 3^2 ,3^3 , 3^4 ..we dont require the 6th weight nor we can have less than 5 weights .
  
  we can obtain..
  1 itself
  3-1=2
  3 itself
  3+1=4
  9-3-1=5
  9-3=6
  9-3+1=7
  9-1=8
  9+1=10
  9+3-1=11
  9+3=12
  9+3+1=13
  27-9-3-1=14
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  27+9+3+1=40
  81-27-9-3-1=41
  81-27-9-3 =42
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  81+27+9+3=120
  So this way we effectively used only 5 weightz viz, 1,3,9,27,81
  
  Now suppose if we had to weigh in such a weigh that we can only add weight and never reduce weight ,
  then ,
  the weights required wud be in the powers of 2
  like
  1,2,4,8,16,32 .....and so on.....
• 12 years agoHelpfull: Yes(1) No(1)