Once upon a time there was a mighty and wise wizard who was
looking for a new student. Because he didn't like the idea
to tell someone stupid the secrets of magic he created the
following test. He took 12 coins which looked all the same.
One of these coins had a slightly different weight. The
student's job was to find out which one was different and to
state if its weight is lighter or heavier than the weight of
the other coins. Therefore the wizard created a magic pair
of scales. The scales could be used three times. After that
the scales disappeared in a blinding flash of light. Using
the scales once means to put the same amount of coins on both
sides, to check the equilibrium and to remove all coins.
Find out how to pass the test without guessing! The complete
solution needs a lot of explanation. To simplify the answer
just sum up the minimum number of coins to be placed on the
scales for the most complicated case, e.g. 5+5 / 3+3 / 2+2 = 20.
Correct answer: 16
Explanation:
The scheme below shows how to solve the problem. The first step is to devide the 12 coins into the three groups A,B and C with 4 coins each.
Note, that you don't know if the different coin is lighter or heavier, so weighting 6+6 at the beginning doesn't help anything. This would be the right start
if you know that the one coin is heavier.
In the scheme below 14 or 16 coins are used (depending on the result). So the most complicated case needs 16.
=====================================================
#1
AAAA | BBBB
1234 5678
left heavier => #6
right heavier => #10
equal => #2
======================================================
#2
C C | C A
9 10 11 1
left heavier => #4
right heavier => #5
equal => #3
======================================================
#3
C | A
12 1
left heavier => C12 heavier (4/2/1) -> 14
right heavier => C12 lighter (4/2/1) -> 14
equal => not possible
======================================================
#4
C | C
9 10
left heavier => C9 heavier (4/2/1) -> 14
right heavier => C10 heavier (4/2/1) -> 14
equal => C11 lighter (4/2/1) -> 14
======================================================
#5
C | C
9 10
left heavier => C10 lighter (4/2/1) -> 14
right heavier => C9 lighter (4/2/1) -> 14
equal => C11 heavier (4/2/1) -> 14
======================================================
#6
AAB | ABC
125 369
left heavier => #8
right heavier => #9
equal => #7
======================================================
#7
B | B
7 8
left heavier => B8 lighter (4/3/1) -> 16
right heavier => B7 lighter (4/3/1) -> 16
equal => A4 heavier (4/3/1) -> 16
======================================================
#8
A | A
1 2
left heavier => A1 heavier (4/3/1) -> 16
right heavier => A2 heavier (4/3/1) -> 16
equal => B6 lighter (4/3/1) -> 16
======================================================
#9
A | C
3 9
left heavier => A3 heavier (4/3/1) -> 16
right heavier => not possible
equal => B5 lighter (4/3/1) -> 16
======================================================
#10
BBA | BAC
561 729
left heavier => #11
right heavier => #12
equal => #13
======================================================
#11
B | B
5 6
left heavier => B5 heavier (4/3/1) -> 16
right heavier => B6 heavier (4/3/1) -> 16
equal => A2 lighter (4/3/1) -> 16
======================================================
#12
B | C
7 9
left heavier => B7 heavier (4/3/1) -> 16
right heavier => not possible
equal => A1 lighter (4/3/1) -> 16
======================================================
#13
A | A
3 4
left heavier => A4 lighter (4/3/1) -> 16
right heavier => A3 lighter (4/3/1) -> 16
equal => B8 heavier (4/3/1) -> 16
======================================================
