  Problem 5A

(We told you this was hard)

This one's a (great) classic. It has been presented in many different ways. At one point, it was known as the Counterfeit Coin Problem: Find a single counterfeit coin among 12 coins, knowing only that the counterfeit coin has a weight which differs from that of a good coin. You are only allowed 3 weighings on a two-pan balance and must also determine if the counterfeit coin is heavy or light.

12 coins, call them ABCDEFGHIJKL:

First weighing: Compare ABCD and EFGH.

• If ABCD=EFGH, you know the special coin is among IJKL. Use your 2nd weighing to compare AI and JK (you know A is ordinary):
• If AI=JK, you know L is the odd coin. You may compare L and A in the 3rd weighing to determine if L is heavy or light.
• If AI and JK are not equal, you know L is ordinary. Use your 3rd weighing to compare J and K.
• If J=K, you know I is the special coin (the result of the second weighing tells you if it's heavy or light).
• Otherwise, the special one is either J or K and you can tell which: It's the heavier one if we had AI<JK on the second weighing, it's the lighter one if we had AI>JK.
• If ABCD is not equal to EFGH, you know IJKL are ordinary (we may use L as a known ordinary coin in the rest of the procedure). Also, you will always know from this first weighing whether the special coin is heavy or light, once you've determined which it is.
Use the 2nd weighing to compare ABE and CFL :
• If ABE=CFL, you know the special coin is among DGH. You may use the 3rd weighing to compare G and H :
• If G=H, then D is the special coin.
• If G and H are not equal, the special coin is the heavier one if we had ABCD<EFGH, and the lighter one otherwise.
• If ABE and CFL are not equal, the special coin is among ABCEF and we may distinguish four cases :
• ABCD>EFGH and ABE>CFL :
Either F is light or one of AB is heavy.
Compare A and B in the 3rd weighing to find out.
• ABCD>EFGH and ABE<CFL :
Either E is light or C is heavy.
Compare C and L in the 3rd weighing to find out.
• ABCD<EFGH and ABE<CFL :
Either F is heavy or one of AB is light.
Compare A and B in the 3rd weighing to find out.
• ABCD<EFGH and ABE>CFL :
Either E is heavy or C is light.
Compare C and L in the 3rd weighing to find out.

The table below summarizes the entire procedure:
First Weighing  Second Weighing  Third Weighing
ABCD = EFGH  AI = JK A = L   is not possible.
A < L  Þ  L is heavy.
A > L  Þ  L is light.
AI < JK J = K  Þ  I is light.
J < K  Þ  K is heavy.
J > K  Þ  J is heavy.
AI > JK J = K  Þ  I is heavy.
J < K  Þ  J is light.
J > K  Þ  K is light. ABCD > EFGH  ABE = CFL G = H  Þ  D is heavy.
G < H  Þ  G is light.
G > H  Þ  H is light.
ABE < CFL C = L  Þ  E is light.
C < L   is not possible.
C > L  Þ  C is heavy.
ABE > CFL A = B  Þ  F is light.
A < B  Þ  B is heavy.
A > B  Þ  A is heavy. ABCD < EFGH  ABE = CFL G = H  Þ  D is light.
G < H  Þ  H is heavy.
G > H  Þ  G is heavy.
ABE < CFL A = B  Þ  F is heavy.
A < B  Þ  A is light.
A > B  Þ  B is light.
ABE > CFL C = L  Þ  E is heavy.
C < L  Þ  C is light.
C > L   is not possible.

Problem 5B

BC signifies "Before Christ".  No one knew Christ was coming, so there would never be a BC label on anything authentic from the period before Christ.  BC is an historical designation used to differentiate the time periods at any point AFTER Christ was born.   