What is the smallest value of $n$ such that the final digit of $13^n$ is one more than the digit adjacent to it? 
What is the smallest value of $n$ such that the final digit of $13^n$ is one more than the digit adjacent to it?

If you have a computer, it is easy to check that the answer to this question is $14$, but I was wondering if there is a way to do it without brute-forcing or using a computer.

I suppose you want to find $n$ such that $13^n$ is of the form $*(x)(x+1)$ for some $x\in\{0,\cdots,9\}$.
This is equivalent to finding $n$ such that $13^n\equiv 11x+1\pmod{100}.$
Then notice that $\phi(100)=\phi(4)\times\phi(25)=2\times20=40,$ so, by Fermat's little theorem, $13^{40}\equiv 1\pmod{100}.$ In fact, $$\begin{cases}13\equiv1\pmod4\\13^{20}=13^{\phi(25)}\equiv1\pmod{25}\end{cases},$$ hence $13^{20}\equiv1\pmod{100}.$ (Furthermore, $13^{10}\equiv13^{2}\times((13^{2})^2)^2\equiv-6\times-4\equiv-1\pmod{25},$ so the order of $13$ modulo $25,$ hence modulo $100,$ is $20.$)
So we are trying to solve $$\begin{cases}1^n\equiv-x+1\pmod4\\13^n\equiv11x+1\pmod{25}\end{cases}.$$
Thus $x$ is divisible by $4,$ say $x=4\times y,$ and we are to solve $13^n\equiv44y+1\equiv-6y+1\pmod{25}, y\in\{0, 1, 2\}.$ So $13^n\equiv1, -5, -11\pmod{25}.$
The first case is answered by $n=20.$
Now notice that $\begin{cases}13^2\equiv-6\\13^3\equiv-3\\13^4\equiv11\\13^{10}\equiv-1\end{cases}\pmod{25}.$ So $13^{14}
\equiv-11\pmod{25}.$ And this solves the problem.
Why is this the minimal one? As the order of $13$ modulo $100$ is $20,$ the first $20$ powers of $13$ cannot repeat itself, and it is easy to see that any power of $13$ cannot be divisible by $5,$ thus $14$ is the minimal one.
Hope this helps.
