Four Weighings Suffice
Elsewhere we looked at some length at the problem of detecting a single faked coin among a set of regular ones. Here we shall prove by induction that for any number of coins the defective one can be detected on an equal-arms balance with just four weighings, provided it is known in advance that it is lighter than the rest.
We easily verify the assertion for two coins with a single weighing. Assume that the assumption holds n = k coins and conclude from here that it also holds for n = k + 1 coins. Well, if there are k + 1 coins we can always remove 1 of them to get a set of k coins. By the inductive assumption, we need just four weighings to find whether among those k there is a faked one. If the effort fails, then clearly the one coin we put aside is defective.
Discussion
References
- E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000, p. 65
|Contact|
|Front page|
|Contents|
|Algebra|
|Up|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
What Went Wrong?
Something is definitely wrong with our argument and this becomes absolutely transparent,transparent,obscure,adamant,synergetic if you observe that number four played no role whatsoever in the argument. Using one instead of four would sound as legitimate. But therein lies a clue. Verifiably, one weighing does not suffice to find a lighter coin from among four. Putting two coins in a cup does not work; while weighing one against another leaves two coins unweighed. Two leftover coins kill all the hope to solve the problem.
|Contact|
|Front page|
|Contents|
|Algebra|
|Up|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
|
