Bob,
I already gave you the solution to the 1st scenario in post #26 & #27.
Here's the solution to the 2nd scenario, where we weighed 1,2,3,4
vs. 9,10,11,12, which results in 'unequal', where either one of the
coins in (1,2,3,4) is 'light' or one of (9,10,11,12) is heavier. In
this example problem #11 is the 'heavy' coin, remember?
To continue...
SCENARIO #2
step-a) 1,2,3,4 vs. 9,10,11,12
result: 'unequal' (we know that 5,6,7,8 are 'good' coins)
step-b) Let's call 1,2,3,4 the 'light' group; 9,10,11,12 'heavy' grp.
What we want to do is move two coins of the 'light' group to
the 'heavy' side; one of the 'heavy' coins to the 'light' side
and put aside two of the 'heavy' coins, leaving 3 coins now
on each side (plus 2 'heavy' coins not used). ( You could also
do this the other way, 2 'light' coins not used, and move 1
light coin to the 'heavy' side, then move 2 'heavy' coins to
the light side.)
i) 1,2,9 vs. 3,4,10
result: 'equal'. This means that the 2 'heavy' coins we put
aside contains the 'odd' coin and is 'heavy' ! Thus,
we just weigh them against each other ( 11 vs. 12),
and of course, we'll see that 11 is 'heavier' than 12!
Let's say however that instead of 11,12 being put aside, we chose
9,10...
ii) 1,2,10 vs. 3,4,11
result: 'unequal'. 1,2,10 is on the 'light' side and 3,4,11
is on the 'heavy' side. What now? We know that coin
10 came from the 'heavy' side and cannot be 'light';
we also know that coins 3,4 came from the 'light' side
and cannot be 'heavy'! Thus, we are left with coins
1,2 (light) and 11 (heavy). So, for the 3rd weighing
just weigh the two coins on the same side 1 vs. 2
which would result in 'equal', leaving 11 as the odd
'heavy' coin ! (If one of them were really 'light'
then that would have been the 'odd' ball).
Dutch Red.
P.S. Somebody gave me this problem a long time ago, and it was
stated differently:
Given 12 coins, all of them of equal weight except one coin,
with using a weighing scale just 3 times, find the odd coin
and tell me whether it's heavier or lighter.
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