Second attempts:
Split into 3 groups of 4, (X, Y, Z)
Scanario 1
(1st weigh)
If X = Y, Z = Anomaly
Split z in 4. z1, z2, z3, z4.
(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. z1 is heavier and the anomaly.
Else = zb 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?
First (failed) attempt:
Split into 3 groups of 4, (X, Y, Z)
Scanario 1
(1st weigh)
If X = Y, Z = Anomaly
Split z in 4. z1, z2, z3, z4.
(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. z1 is heavier and the anomaly.
Else = zb 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?
First (failed) attempt:
.
