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You are an intrepid explorer who has been caught in a trap by a primitive tribe. They lock you in a room with no exit but one door. The door has a pressure sensitive release - incorrect pressure ($>0g$) will cause immediate death. There are 10 piles each of 20 heavy stone discs (200 in total). You know what each disc should weigh (I can tell you that the number is in whole grams).

You have some facts:

  • You know that one of the piles is made of discs that are missing 15.5 grams of the normal weight of a disc.
  • You also know that another distinct pile is made of discs that are missing 6.5 grams of the normal weight.
  • You also know that all the discs in all the other piles weigh exactly the right amount.

You have weighing scale (non-balancing) that is accurate to the gram.

The tribe leader has set the rules. You must determine which piles are normal, which pile is off by 15.5 grams and which is off by 6.5 grams using the scale only once. Only by putting one disc from each of the 2 piles which are not normal on the pressure pad to release the door will you survive.

Gamow
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d'alar'cop
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  • What is "exactly the right amount" or the total weight of a normal pile? – warspyking Oct 08 '14 at 23:11
  • Not telling.... – d'alar'cop Oct 08 '14 at 23:12
  • Is this a duplicate of Coin weighing problem? –  Oct 09 '14 at 00:06
  • @Emrakul It appears to be a slightly more complicated variant – d'alar'cop Oct 09 '14 at 00:14
  • Is the scale a balancing scale where you put weights on both side, or does it just tell you the weight directly? – Ben Aaronson Oct 09 '14 at 00:17
  • It's not a balancing scale. It tells you the weight directly. – d'alar'cop Oct 09 '14 at 00:18
  • For "accurate to the gram"- does it consistently round up or down half values? So could 6.5 show up as either 6 or 7, or is it consistently one or the other? – Ben Aaronson Oct 09 '14 at 00:27
  • @BenAaronson It consistently rounds down, sir – d'alar'cop Oct 09 '14 at 00:28
  • I'd think Joe's answer would work then. I just looked and I can't find any sums of multiples of 6.5 and 15.5 <= 10 which are within 0.5 of each other. – Ben Aaronson Oct 09 '14 at 00:30
  • @BenAaronson You are right. But the answer needs to take all that into account. Although indeed the question should choose the weights more carefully. – d'alar'cop Oct 09 '14 at 00:37
  • Would be interesting to tweak the numbers to create a version of this where exactly one of the sums of multiples was too close. Then you'd have to make the extra connection that you could avoid one of the pile sizes by weighing no discs at all from a pile – Ben Aaronson Oct 09 '14 at 00:39
  • @BenAaronson Yes, that was the intention actually... it wouldn't be that hard. There must be one or more solutions for the 2 off weights, $i$ and $j$, such that $\lfloor xi \rfloor = \lfloor yj \rfloor \land x \neq y \land 0 < x,y \leq 20$ to cause the ambiguity. And then depending on the method of resolving the situation there would need to be other conditions. I had in mind: no solution for $i$ and $j$, such that $2xi = 2yj \land x \neq y \land 0 < x,y \leq 10$ – d'alar'cop Oct 09 '14 at 00:52
  • I was actually looking for a catch like this, and got confused when I didn't find one xD I assumed I was missing the entire point :-P – Joe Oct 09 '14 at 01:00
  • Seems like a spinoff of the sugar cube riddle we had recently. Also, the answer is very much similar. – Tim Couwelier Oct 09 '14 at 12:17
  • http://puzzling.stackexchange.com/questions/2184/find-the-different-sugar-cubes , and even that is said to be a duplicate of the coin weighing problem, as referenced above. – Tim Couwelier Oct 09 '14 at 12:19
  • @TimCouwelier It's a variation I suppose. Personally I feel that it's different enough to justify its existence. e.g. the numerical complications are greater. I think of it as chess positions being similar but not the same... or knight/knave puzzles being variations of each other (but all interesting) – d'alar'cop Oct 09 '14 at 12:20
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    Hence me listing it as a spinoff, not a duplicate :) – Tim Couwelier Oct 09 '14 at 12:20
  • @TimCouwelier Yes, I noticed. Thanks Tim ;) – d'alar'cop Oct 09 '14 at 12:21

1 Answers1

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Since we know what each disc should weigh ($n$), we can weigh 1 disc from pile 1, 2 from pile 2, etc (10+9+8... = $55n$). However, since some weights are half-grams, we'll need to double up instead (20+18+16... = $110n$).

Subtract that from the actual weight we read on the scales, then work out how to make the remainder from 13s and 31s. There is no multiple of 13 and 31 which can cause any problems with the math.

This will tell us which piles the abnormal discs came from, so finally just take 1 from each of those piles to put on to the pressure plate.

Joe
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