Solution to the Heavy Coin puzzle

[This is a solution to the heavy coin puzzle posted earlier]

A brief description of the problem:

You have $96$ coins of weight $H$, $4$ of weight $L$, with $H \gt L$. With two or fewer uses of a two pan balance, find a heavy coin.

Solution

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Make three groups, $A$ with $33$ coins, $B$ with $33$ coins and C with $34$ coins.

Weigh $A$ against $B$.

If $A$ is heavier, then $A$ can have at most one light ($L$) coin.

Now we can take two coins from $A$ and weigh them against each other. Pick the heavier one if of different weight, or pick any one if equal.

The case when $B$ is heavier is similar to the above case.

If $A$ and $B$ are of equal weight, then we have the following possible scenarios

$$\begin{array}{|c|c|c|} \hline A&B&C\\ \hline 33H&33H&30H+4L\\ \hline 32H+L&32H+L&32H+2L\\ \hline 31H+2L&31H+2L&34H \\ \hline \end{array}$$


Now move one coin from $A$ to $B$ (call the result as $B_a$).

Weigh $B_a$ against $C$.

If $B_a$ is heavier than $C$, then the coin you moved was a heavy coin.

If $C$ is heavier, then any coin in $C$ is a heavy coin.

If $B_a$ and $C$ are equal, then the coin you just moved is a light coin, and the $32$ coins left in $A$ are all heavy.