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$$(43)_x=(y3)_8$$

Since a number in base$-k$ can only have digits from $0$ to $(k-1)$, we can conclude that: $x \geq 5$ and $y \leq 7$

Now, the original equation, when converted to decimal base gives:

$$\begin{align}4x^1 + 3x^0 &= y(8^1)+3(8^0)\\[0.8em]4x + 3 &= 8y + 3\\[0.8em]x&=2y\end{align}$$

So, we have the following constraints:

$$x\geq 5\\y \leq 7\\x=2y\\x,y \text{ are integers}$$

The set of values of $(x,y)$ that satisfy these constraints are:

$$\begin{align}(x&,y)\\\hline\\(6&,3)\\(8&,4)\\(10&,5)\\(12&,6)\\(14&,7)\end{align}$$

**I am counting 5 pairs of values.**

+8 votes

(43)_{x }=(y3)_{8}

From this equation it is clear x should be greater than 4 and y should be less than 8.

x>=5, and y<=7

Now convert this equation in decimal

_{=> 4*x+3=y*8+3}

_{4x=8y}

_{ie x=2y }

With given constraint (x>=5 && y<=7)

possible values of x and y are (6,3), (8,4),(10,5),(12,6), (14,7)

So possible number of solution is 5.

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