A set of 101 coins consists of 100 genuine coins of equal weight plus one fake coin which is either
a. lighter
OR
b. heavier
than a genuine one.
Using only equal-arm-balance, please determine in two weighings whether it is a. or b.
IMHO it is a nice puzzle, but I doubt whether the result has any practical meaning.
Objections? Let my know.
Source: Leningrad Math. Competition, former USSR.
Put 30 on one side and 30 on the other (could have used 26 through 33 on each side, but 30 is a nice round number).
If they weigh equally, you now have 60 known to be good. Weigh 41 known good ones against the 41 yet to be weighed. If the previously unweighed now go down you know the bad one is heavier than the rest; if the previously unweighed now go up, the bad one is lighter.
If the first weighing was unbalanced,then weigh, say, the lighter side against 30 of the 41 previously unweighed, since those are known to be good. If they weigh equally, the unknown coin must be heavier as it was part of the original heavy side; if the lighter side from the first weighing is still the lighter side in the second weighing, the know the bad coin is lighter and part of this group. A similar method would be if for some reason you chose the second weighing to be the heavier side against the unweighed coins assumed to be good.
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Posted by Charlie
on 2022-04-18 11:43:54 |
solution
