Two prisoners, 64 coins and a escaque secreto

1. Number the 64 boxes from 0 to 63.
2. Think of each number as a 6-bit string.
3. Usa the xor of the indices of the coins that muestran head.

Answer: yes. They can use a special binary addition, called xor, to convert the entire board into a number from 0 to 63. First they number the boxes from 0 to 63. Before entering, they agree that a heads coin counts and a tails coin does not count. The first prisoner looks at all the coins that are heads and calculates the xor of their indices. Call it $S$. If the secret box has number $T$, the first prisoner must flip the coin from the box: $$
S\oplus T.
$$ Why does it work? When you flip a coin, its index enters or exits the total xor. In both cases, the new xor of the board becomes exactly $T$. Then the second prisoner enters, calculates the xor of the indices of the coins that are heads and obtains $T$. That number tells you which was the secret box.