What is the remainder?

We need to find the following.

$\begin{aligned} 12^{\color{#3D99F6}13^{14}} & \equiv 12^{\color{#3D99F6}2n+1} \text{ (mod 145)} & \small \color{#3D99F6} \text{where }n \in \mathbb N \text{, since }13^{14} \text{ is odd.} \\ & \equiv (145-1)^n \cdot 12 \text{ (mod 145)} \\ & \equiv (-1)^n \cdot 12 \text{ (mod 145)} & \small \color{#3D99F6} \text{Since }n \text{ is even (see note)} \\ & \equiv \boxed{12} \text{ (mod 145)} \end{aligned}$

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Note: $2n+1 = 13^{14}$ , $\implies 2n = 13^{14} - 1 = (13^7-1)(13^7 + 1)$ . Note that both factors $13^7-1$ and $13^7 + 1$ are even, hence $2n$ is a multiple of 4 and $n$ is even.