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We have 5 boxes with 20 coins each. And coins of three boxes weigh 10, one of them weighs 9 and the other weighs 11. How can we differentiate between them by using a scale(digital, not the one with two cups) only once, with 1 unit accuracy?

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    $\begingroup$ Weigh them? shrugs This puzzle needs more restrictions to be meaningful. $\endgroup$ – mr23ceec Dec 11 '17 at 16:14
  • $\begingroup$ @mr23ceec "using a scale only once" $\endgroup$ – Apep Dec 11 '17 at 16:15
  • $\begingroup$ Is it a balancing scale (two cups and one side is equal to the other) or a spring scale (one cup and it gives you a numerical answer, usually in a certain range) $\endgroup$ – mr23ceec Dec 11 '17 at 16:23
  • $\begingroup$ One cup digital scale $\endgroup$ – Nemexia Dec 11 '17 at 17:11
  • $\begingroup$ each coin weigh 10 or whole box weigh 10? $\endgroup$ – Oray Dec 11 '17 at 18:07
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Place 1 coin from the first bag on the scale 2 coins from the second bag, 4 coins from the third bag, 8 coins from the fourth bag and 16 coins from the fifth bag

subtract 310 from the total, and the result will tell you which boxes contain the wrong coins. (it is I - J, where I and J are 2^n and n is the bag number)

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  • $\begingroup$ So its kinda binary code? $\endgroup$ – Nemexia Dec 11 '17 at 16:54
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Assuming each coin weigh 10 for 3 boxes, 9 for another box, 11 for the other one.

This is pretty straight forward question. Just choose $0,1,3,7,12$ and this is also the minimum number of coins you can choose as well. You will get different $20$ results except the boxes with coins with the same weight coin.

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  • $\begingroup$ Minimum number of coins if you take out and weigh coins. If you left off one box completely then weighed 4 boxes with coins removed from each box? I think that does it handling fewer individual coins. $\endgroup$ – Bob Dec 11 '17 at 19:01

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