Question 45

Given the equation:

In base $x$, the digits are 4 and 3. For these digits to be valid, the base must be strictly greater than the maximum digit:

In base 8, the digits are $y$ and 3. For $y$ to be a valid non-zero leading digit:

Convert both numbers to base 10:

Substitute each possible integer value of $y \in \{1, 2, 3, 4, 5, 6, 7\}$ and test the condition $x \ge 5$:

• $y = 1 \implies x = 2 < 5$ (invalid)
• $y = 2 \implies x = 4 < 5$ (invalid)
• $y = 3 \implies x = 6 \ge 5$ (valid solution: $(6, 3)$)
• $y = 4 \implies x = 8 \ge 5$ (valid solution: $(8, 4)$)
• $y = 5 \implies x = 10 \ge 5$ (valid solution: $(10, 5)$)
• $y = 6 \implies x = 12 \ge 5$ (valid solution: $(12, 6)$)
• $y = 7 \implies x = 14 \ge 5$ (valid solution: $(14, 7)$)

There are exactly 5 possible pairs of $(x, y)$ that satisfy all constraints.