Consider the equation $(43)_x = (y3)_8$ where $x$ and $y$ are unknown. The number of possible solutions is _________

Convert both sides of the equation $(43)_x = (y3)_8$ to base 10:

Subtracting 3 from both sides gives:

Now, we determine the valid constraints on the digits and bases:

1. In base $x$, the digits are $4$ and $3$, which requires base $x > 4$.

2. In base 8, the digits are $y$ and $3$, so $0 \le y \le 7$. Since $y$ is the most significant digit, $y \ge 1$.

3. Since $x = 2y$ and $x > 4$, we have $2y > 4 \implies y > 2$.

Therefore, the possible integer values for $y$ are:

For each value of $y$, the corresponding $x = 2y$ is:

• $y = 3 \implies x = 6$ (Valid: $(43)_6 = (33)_8 = 27$)

• $y = 4 \implies x = 8$ (Valid: $(43)_8 = (43)_8 = 35$)

• $y = 5 \implies x = 10$ (Valid: $(43)_{10} = (53)_8 = 43$)

• $y = 6 \implies x = 12$ (Valid: $(43)_{12} = (63)_8 = 51$)

• $y = 7 \implies x = 14$ (Valid: $(43)_{14} = (73)_8 = 59$)

Thus, there are 5 possible solutions.