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.
 A: Note that since $$13^{1} \equiv 1 \pmod 4$$ and that $$13^{20} \equiv 1 \pmod {25}$$ We have that $$13^{20} \equiv 1 \pmod {100}$$ So we merely need to check $n \le 20 $. 
Now, proceed to note that $13^{n} \equiv 1,3,9,7 \pmod {10}$, we are looking for solutions $$13^{n} \equiv 1,23,89,67 \pmod {100}$$Calculations are not avoidable in this case, I'm afraid. There may be several manipulations to make this easier, but there are (I think) none simple for this case. 
One thing to make it simpler would be to note that $$ord_{100}(67)=ord_{100}(23)=20$$, so such $n$ that $$13^{n} \equiv 23,67 \pmod {100}$$ would be coprime to $20$. 
A: 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.
