bases trouble be me lot

You can easily translate from any base to base ten using the following equation:

$d_n \cdot b^n + d_{n-1} \cdot b^{n-1} + \ldots + d_1 \cdot b^1 + d_0 \cdot b^0$

Where $b$ is the base, $d_n$ is the digit in the $n$ 's place, and $n$ is the number of digits in the number minus one.

(For example, 1303 in base 3 is $1 \cdot 3^3 + 3 \cdot 3^2 + 0 \cdot 3^1 + 3 \cdot 3^0 = 57$ )

So, if $34$ in base $x$ equals $43$ in base $y$ , this translates to a very simple linear equation:

$3x + 4 = 4y+3 \Rightarrow y = \frac {3x + 1}{4}$

The solutions must be positive integers, but they must also be greater than any digit in the problem (or the number wouldn't even be valid for that base). The first two sets of integer solutions to this equation, $(1, 1)$ and $(5, 4)$ , do not satisfy this condition. The first valid solution is $(9, 7)$ .

So the answer is $9 + 7 = 16$

[This is not a solution.]