2
$\begingroup$

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.

Improve this question
$\endgroup$
18
  • $\begingroup$ What is "exactly the right amount" or the total weight of a normal pile? $\endgroup$
    – warspyking
    Oct 8, 2014 at 23:11
  • $\begingroup$ Not telling.... $\endgroup$
    – d'alar'cop
    Oct 8, 2014 at 23:12
  • $\begingroup$ Is this a duplicate of Coin weighing problem? $\endgroup$
    – user20
    Oct 9, 2014 at 0:06
  • $\begingroup$ @Emrakul It appears to be a slightly more complicated variant $\endgroup$
    – d'alar'cop
    Oct 9, 2014 at 0:14
  • 1
    $\begingroup$ Hence me listing it as a spinoff, not a duplicate :) $\endgroup$
    – Tim Couwelier
    Oct 9, 2014 at 12:20

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Nice, but this won't work because scale is correct to the gram... +1 for now $\endgroup$
    – d'alar'cop
    Oct 9, 2014 at 0:13
  • $\begingroup$ Ah ok, so it's going to be something like double the numbers, instead (2, 4, 6...) - I'll take a think when I get up :-P $\endgroup$
    – Joe
    Oct 9, 2014 at 0:59
  • $\begingroup$ lol. good job mate. (but do please complete the answer) $\endgroup$
    – d'alar'cop
    Oct 9, 2014 at 1:00
  • $\begingroup$ @d'alar'cop updated the answer for you :-) $\endgroup$
    – Joe
    Oct 9, 2014 at 8:15
  • $\begingroup$ well done :) exactly what I had in mind. $\endgroup$
    – d'alar'cop
    Oct 9, 2014 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.