For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{16} \cdot 2a + 10 \ \land d \ \vert \ 3^{17} \cdot a + 66$
$\Rightarrow d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 3 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 2$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 30 \ \land \ d \ \vert \ 3^{16} \cdot 6a + 132$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 132 - (3^{16} \cdot 6a + 30) \ = \ 102 \ = 
 \ 2 \cdot 3 \cdot 17$
Also, $d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 33 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 5$
$\Rightarrow d \ \vert \ 3^{17} \cdot 17a$ (almost with the same method as before)
So I get $d \ \vert \ 102 \ \land d \ \vert \ 3^{17} \cdot 17a$
After this, I can't see how to continue.

Hint $\ \ \ (d,102) = (d,\,2\cdot 3\cdot 17) = (d,2)\,(d,3)\,(d,17).\, $ Computing these gcds by Euclid & Fermat
$\qquad\quad\ \begin{align} (d,2) &= (\color{#c00}{3^{16}(2a)\! +\! 10}, \ \color{#0a0}{3^{17} a\! +\! 66},\,2) = (\color{#c00}0,\color{#0a0}a,2) = (a,2)\\[.2em]
(d,3) &= (\color{#c00}1,\ \color{#0a0}0,\ 3) = 1,\ \text{   and, finally, using $\,\color{#c00}{3^{16}}\equiv 1\!\!\!\!\pmod{\!17}\ $ we have}\\[.2em]
(d,17) &= (\color{#c00}{2a\!+\!10},\,\color{#0a0}{3a\!-\!2},17) = (2a\!+\!10,a\!+\!5,17) = (0,a\!+\!5,17) 
\end{align}$
Hence $\, (d,102) = (a,2)\,(a\!+\!5,17)$

Remark $\, $ The elimination that you employed to deduce that $\,d\mid 102\,$ is a special case of the determinant criterion presented here. Namely, applying that with $\,b=3^{16}\,$ yields
$\ (ab,1)\mapsto (2ab\!+\!10,3ab\!+\!66)\,$ has $\,\det = 2(66)\!-\!10(3) = 102\ $ so $\ d\mid 102(ab,1) = 102$ 
