Tuesday Puzzle

A rich man hires a laborer for a week. The rich man has a bar of gold and he has to pay the laborer in gold at the end of each day. He has to give him exactly what should be due to him at the time of payment – not more, not less.
What are the minimum number of cuts that need to be made in the bar of gold so that the rich man can pay the laborer fairly each day ?

9 Comments:

  2. Leena
    September 29, 2006 at 12:04 pm

    6 will be the max no of cuts.
    HINT – If you know bits & bytes, you may be able to solve this easily.

  3. Leena
    September 29, 2006 at 12:06 pm

    Another HINT : The cuts dont have to make equal lengths of the pieces.

  5. Leena
    September 29, 2006 at 12:32 pm

    Nope. Another hint : You may be able to take things back.

  6. BhaveshG
    September 29, 2006 at 12:33 pm

    Week is defined as a full Week (7 days) or a Typical US working week (5 days) ??? just kidding…

  7. prashant
    September 29, 2006 at 1:08 pm

    3 cuts. They will be of inequal widths.
    1 slab will be equal to 1 day’s of work.
    2nd slab will be equal to 2 days of work.
    3rd slab will be equal to 4 days of work.

    On 1st day –> 1st slab.
    On 2nd day –> take back the slab given earlier and give second slab.
    On 3rd day –> give 1st slab and dont take back anything.
    On 4th day –> take back both slabs and give the 3rd slab (worth 4 days of work)
    On 5th day –> give 1st slab also
    on 6th day –> take 1st slab and give 2nd slab
    on 7th day –> give 1st slab again

  8. Leena
    September 29, 2006 at 1:34 pm

    Good job !!! The reasoning is correct but the answer is not. Since its 3 pieces, you need only 2 cuts.
    This puzzle is based on the binary system where 3 bits can represent 0 to (2^3)-1 decimal numbers.
    000 = 0
    001 = 1
    010 = 2
    011 = 3
    100 = 4
    101 = 5
    110 = 6
    111 = 7

  9. prashant
    September 29, 2006 at 1:40 pm

    Do i get a prize for the reasoning at least. 😀

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