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Weighing Up the Competition

Posted by Karl Sharman on May 22, 2011 – 8:07 am

Chris has just started a new company selling weighing instruments – “Chris-Mass”. (Geddit?!?! Christmas – Chris-Mass, you know – weight!?!? – sometimes I excel myself!)
His first order has just been shipped to you. It comprises of a super-accurate beam balance and a 40 lb. rock, this system will allow you to accurately weigh items from one to forty pounds, and every integer value in between. Some dis-assembly required, of course.
Obviously the forty pound rock will have to be cut up with a high-powered diamond-tipped saw (not included), because as it is, it will only let you accurately weigh other forty pound rocks.
What is the fewest number of pieces the rock will need to be cut into to measure any item weighing an integer amount from one to forty pounds, inclusive (assuming you can manage to get the rock into the precise fractions you want)?

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17 Responds so far- Add one»

  1. 1. SP Said:

    I came up with 5.

  2. 2. Chris Said:

    4 pieces.

  3. 3. Andy Said:

    I agree with five. 2, 3, 5, 10 & 20.

  4. 4. Chris Said:

    The 5 weights are wasteful. That set provides multiple different ways of weighing several of the weights.

    It can be done with only 4 weights.

  5. 5. Nathan Said:

    I agree with chris. 1 3 9 and 27 lb weights will measure any integer from 1 to 40 inclusive.

  6. 6. Chris Said:

    That’s the boyos. There’s a variation on this puzzle. How many weights do you need to weigh out objects that are known to be of exact integer weight from 1 to 81?

  7. 7. Nathan Said:

    This too can be done with four weights.

  8. 8. Chris Said:

    Hi Nathan. That’s right :)

  9. 9. cazayoux Said:

    Logically, I can understand 4 stones for 81 different measurements.

    N = az * by * cx * dw
    z,y,x,w are the stones
    a,b,c,d each have three different values [0,+1,-1]
    With each having 3 values there are 3*3*3*3 = 81 options

    With stones 1,3,9,27 I can also see that we can actually measure -40 to 40 lbs.
    Still trying to figure out how to shift that range from [-41..40] to [1..81].

  10. 10. cazayoux Said:

    Can I move the fulcrum?!!

  11. 11. cazayoux Said:

    If I put all 40 pounds on one side, I’d move the fulcrum until balanced (x = 0 lbs).
    However, this gives me a range that I can weigh from 0 to 80 pounds, not 1 to 81.

    hmmmmmmmmm.

  12. 12. Nathan Said:

    Here is a hint:
    If you do a weighing and the object weights more then what you put on, and if you do the next highest weight and the object weighs less, it is possible to determine what the weight of the object is.

  13. 13. Chris Said:

    … and another. It doesn’t have to be 40 lbs, and another, none of the items you are weighing is greater than 81 lbs or less than 1 lb. (So you only have to distinguish among 80 distinct weights that are guaranteed to be one of 1,2,3,…81).

  14. 14. Karl Sharman Said:

    The rock will suffice in as few as four pieces.

    The trick in solving the puzzle is realizing that rock framents can be placed on either plate, coupled with what you’re trying to weigh. If you have, say, a one pound chunk and a three pound chunk, you can weigh a two pound item. Simply weigh the item AND the one pound rock against the three pounder. If they balance, it’s two pounds. Having established this, the next problem is deciding on the rock sizes which allow all integer combinations.
    The correct sizes are: 1, 3, 9, and 27 pounds

    Jumping beyond Chris’s 81 lb problem, a 121 pound weight broken into 5 pieces will give you every integer up to 121. You’ll need 1, 3, 9, 27, and 81. Is this a coincidence?

  15. 15. DP Said:

    Chris: is it 2, 6, 18, and 54? You don’t need to weigh an item that is say 1 lb. Since you know it is an integer weight, and lighter than 2 lbs, it must be 1. for 2 lbs, you have that weight.
    For 3 lbs try the 2…it’s heavier you say? Well then try weighting it with the 2 against the 6 as if it is 4 lbs…it’s lighter you say? Then it must be 3 lbs!
    And so on.

  16. 16. Chris Said:

    Hi Karl. Even without proving it, it’s pretty obvious than eah weight is 3 times the previous one.

    Hi DP. That’s the fellas and the reason :)

  17. 17. Karl Sharman Said:

    Hi Chris – I thought it was a nice, possibly over-obvious clue…

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