In a pile, there are 11 coins: 10 coins of common weight and one coin of different weight (lighter or heavier). They all look similar.

Using only a balance beam for only three times, show how you can determine the 'odd' coin.

Open problem (i cannot solve this myself): how many more coins (with the same weight as the ten) can we add to that pile so that three weighing still suffices? My conjecture is zero, though my friend guessed that adding one is possible. The best bound we can agree upon is < 2.

(2) A and B differ in weight:
Call the lightest group A and the heaviest group B. Now you know that the remaining group C consists of correct coins. Then do the following weighing: take two coins from group A and two coins from group B (A1, A2, B1, and B2) and weigh them against one coin from A, one coin from B and two coins from C (A3, B3, C1, and C2). Now the following three situations are possible:
(2a) A1 + A2 + B1 + B2 are as heavy as A3 + B3 + C1 + C2:
This means that A4 or B4 has a different weight, now you can determine which one it is by weighing for instance A4 against C1 (= a correct coin).
(2b) A1 + A2 + B1 + B2 are lighter than A3 + B3 + C1 + C2:
This means that either A1 or A2 are of different weight (lighter) or B3 is of different weight (heavier). Now weigh A1 + B3 against C1 + C2 which can result in the following situations:
(2bi) A1 + B3 are as heavy as C1 + C2:
Which means that A2 has a different weight.
(2bii) A1 + B3 are lighter than C1 + C2:
Which means that A1 has a different weight.
(2biii) A1 + B3 are heavier than C1 + C2:
Which means that B3 has a different weight.
(2c) A1 + A2 + B1 + B2 are heavier than A3 + B3 + C1 + C2:
Similar explanation as in situation 2b...

Posted by Half-Mad on 2002-05-16 02:20:02