There are 10 Boxes, each contains 1000 sugar cubes. The sugar cubes in 9 of the boxes are 10 grams each and the other box has 9 gram cubes.

There is a digital scale that you can use it ONCE. (i.e. put something on the scale, turn it on and read the number)

Find the box with the 9 gram cubes.

Source: Fekraneh.ir

  • $\begingroup$ You should clarify what you mean by using the digital scale once. Does it mean that you can only look at one number that is produced by the scale? (There is a sneaky solution where you can put one of each cube on the scale and then remove them one by one) $\endgroup$ Commented Sep 3, 2014 at 16:36
  • 4
    $\begingroup$ Essentially a Duplicate: puzzling.stackexchange.com/questions/1965/coin-weighing-problem $\endgroup$ Commented Sep 3, 2014 at 17:26
  • $\begingroup$ Treesrule14 , I have edited the description. thanks $\endgroup$
    – Rafe
    Commented Sep 3, 2014 at 19:53

1 Answer 1

  • Set number from 1 to 10 to each box.

  • pick cubes from from boxes with the equal amount of number of each box, and put them on scale.

  • read the last digit of shown number on the digital scale, and subtract this amount from 10. (e.g. weight=543 grams => 10 - 3 = 7 )

    The signle digit result will lead us to the 9 gram cubes box

  • 2
    $\begingroup$ I think you tried to oversimplify. If you take 1 cube from box 1, 2 from box 2 etc, you'll end up with 55 cubes which would be 550gr if all were 10g. The difference is how many 9g cubes were in there, giving the number of the box. It's not always a single digit result as it could be box 10. (though 10 - 0 = 10 is valid, it's not single digit). I'd just say 550 - [scale reading] = box with the 9 gram cubes. $\endgroup$ Commented Sep 3, 2014 at 7:55
  • $\begingroup$ No @TimCouwelier! I dont agree with you. The last digit of 540 would be 0 (not 10), so the result would be: 10 - 0 = 10 .... and it will show that the 9gram cubes are in box number 10. But as you said "550 - [scale reading] = box" is easier to understand. Thanks! $\endgroup$
    – Rafe
    Commented Sep 3, 2014 at 9:03
  • $\begingroup$ Indeed, the logic is valid, it's just that the '10' answer isn't a single digit. But that's nitpicking. Hence just a comment and no downvote $\endgroup$ Commented Sep 3, 2014 at 12:40
  • 1
    $\begingroup$ You could also number the boxes from 0 to 9. Less cubes. $\endgroup$
    – Florian F
    Commented Sep 3, 2014 at 14:13

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