Mathematical_Recreations-Kraitchik-2e

THE FALSE COIN. 825
Hence we can identify the group of 3 coins which contains
the false one. If a, b, c are the 3 coins of that group
we compare in a second balancing the coins a and b.
If a>b, a is the false coin
a<b b" " " "
a=b c" " " "
The solution is similar if the false coin is lighter than
the other.
In the case of N = 3" coins we need n balancings in
order to identify the false coin. The same number of balancings
is required if 3- 1 <N <3".
3. We have 12 coins and we are asked to identify the
false coin and find its defect by 3 balancings.
We distribute the 12 coins, A, B, C ... in 3 groups:
ABCD, EFGH. IJKL.
First
balancing
Second
balancing
Third
balancing
Conclusion
ABCD>EFGH ABE>CDF A>B A>O
A<B B>O
A=B F<O
ABE<CDF C>D C>O
C<D D>O
C=D E<O
ABE=CDF G>H H<O
G<H G<O
ABCD<EFGH ABE>CDF C>D D<O
C<D C<O
C=D E>O
ABE<CDF A>B B<O
A<B A<O
A=B F>O
ABE=CDF G>H G>O
G<H H>O
ABCD=EFGH. The false coin is in the third group.
That is the problem l;which we achieve by 2 balancings.
With N = Y2 (3" - 3) coins n balancings are needed.