Remainder? Part-4

Similar solution as @Tom Engelsman's

$\begin{aligned} N & = 3^{4n-2} + 2^{6n-3} + 1 \\ & = 3^{2(2n-1)} + 2^{3(2n-1)} + 1 \\ & = \blue{9^{2n-1} + 8^{2n-1}} + 1 & \small \blue{\text{Since }2n-1 \text{ is odd}} \\ & = \blue{(9+8) \sum_{k=1}^{2n-2} (-1)^k 9^k \cdot 8^{2n-2-k}} + 1 \\ & = 17 \sum_{k=0}^{2n-2} (-1)^k 9^k \cdot 8^{2n-2-k} + 1 \end{aligned}$

Therefore, $N \bmod 17 = \boxed 1$ .

Let us rewrite the above expression according to:

$3^{4n-2} + 2^{6n-3} + 1 = (3^2)^{2n-1} + (2^3)^{2n-1} + 1 = 9^{2n-1} + 8^{2n-1} + 1$ (i)

Since we have $8$ and $9$ both being raised to an odd integer power in (i), we can factor according to:

$(9+8) \cdot [\Sigma_{k=0}^{2n-2} (-1)^k \cdot 9^{2n-2-k} \cdot 8^{k}] + 1$ (ii)

Since $17 | (9+8) \cdot [\Sigma_{k=0}^{2n-2} (-1)^k \cdot 9^{2n-2-k} \cdot 8^{k}]$ in (ii), the remainder is just $\boxed{1}.$