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You have a bar of vibranium and the Cap needs to use it (not used up or consumed) every day but he only needs 1/7th on the first day, 2/7th on the second day and so on for 7 days.

You have a vibranium knife that can cut through it for the Cap; however the knife is only good to make 2 cuts. You are very good and can cut perfectly.

How can you cut the vibranium bar into 3 pieces so that the Cap can get the perfect amount of vibranium for each day from 1 to 7?


marked as duplicate by klm123, wbogacz, Peter Taylor, mdc32, kaine Oct 28 '14 at 14:06

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    $\begingroup$ very close but perhaps not a dupe: puzzling.stackexchange.com/questions/2191/… $\endgroup$ – Kate Gregory Oct 27 '14 at 20:46
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    $\begingroup$ Logic is the same $\endgroup$ – Huangism Oct 27 '14 at 20:47
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    $\begingroup$ Logic is the same, though I find the chain link version nicer, due to it only requiring a single 'cut' - the nature of the chain link adds to it. $\endgroup$ – Tim Couwelier Oct 28 '14 at 13:01
  • $\begingroup$ Wait, are we talking 7ths of the total bar here or 7ths of what remains on that day? $\endgroup$ – Weckar E. Nov 16 '16 at 9:00

Make two cuts, leaving pieces of $1/7, 2/7, 4/7$ You can select pieces to get the amount required for each day.

  • $\begingroup$ Indeed, did you figure it out or seen this type before? $\endgroup$ – Huangism Oct 27 '14 at 20:43
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    $\begingroup$ There is an old classic of paying for a hotel room with one link per day from a $63$ link gold chain. How few links can you cut so you can always make change. Any time I see a number $2^n-1$ I give binary a try. $\endgroup$ – Ross Millikan Oct 27 '14 at 20:45

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