The 12 Coins Puzzle

[theVOID]

I was in the process of typing it all out when you all responded, i got it done on paper in doodles in a minuet or so, took a long time to make sense of my abstract imagery.

Second attempts:

Split into 3 groups of 4, (X, Y, Z)


Scanario 1

(1st weigh)
If X = Y, Z = Anomaly

Split z in 4. za, zb, zc, zd.

(2nd weigh)
Weigh Za and Zb


a) If za = zb, discard za and zc.
b) If za ≠ zb, discard zb and zd

We now have 2 groups with 1 coin each. za and zb, with weighted groups for each scenario

outcome A = zc is heavier and zd is lighter.
outcome B = za is heavier and zb is lighter.

(3rd weigh)

A) Weigh zb and zd.

If zb = zd. za is heavier and the anomaly.
Else = zc is the anomaly

B) Weigh za and zc

If za = zc. Zb is the anomaly
Else = za is the anomaly.


Scanario 2
(remember we have 3 groups of 4, X, Y and Z.)

(first weigh)
If X = Y, Z = anomaly group (shown above).

If X = Heavier (assuming for this example, just switch heavy with light here on in for the rest of the solution if otherwise), break X into 1,2,3,4 and Y into 5,6,7,8.

Separate into groups of 3, remember what group they were in but mix them up.

Group A = 156
Group B = 278
Group C = 34

(second weigh)

If A = B, Group C is anomaly group (solve this now by weighing one remaining coin against one known average coin, remember 3 and 4 were in the heavy group in this scenario. If scale is even the other coin is the anomaly, if 7 is light or heavy again it is the anomaly).

If A(156) = heavier and B = 278 is lighter, and since we already know that group A, which 1 is in, was the heavy side last time, then either i)1 is the Heavy coin or ii)7 or 8 are the light coin.

(Third weigh)

i) Weigh 7 against 8.

If 7 = 8, 1 is the anomaly
if 7 = heavier, 8 is the anomaly
If 8 = heavier then 7 is the anomaly


Is this correct Adrian?

I hope it is... I've been racking my brain over this one, haven't done any real algebra in quite some time.

Um, I think you're measuring them wayyy more than 3 times.

You can only weigh the coins 3 times, switching coins or groups counts as a weigh in.

You can remove all the coins from the scales, change the stacks and then weigh them again.

The other reason why it seemed like so many is because i have to account for all potential scenarios as there is an element of chance regarding where the anomaly is amongst the coins you weigh at any given moment.