How do I prove this property of the sequence? Given a sequence defined by the recursive formula:
$$x_{1}=1
$$
$$x_{k+1}=x_{k}+(x_{k} \bmod 10)$$
Prove that : $$\forall n \in \mathbb{N}, n>0 \; \exists k (x_{k}=4^{n})$$
All I have tried is that proving this sequence of $x_{k} \bmod 10 $ will enter the cycle $2\rightarrow 4\rightarrow 8\rightarrow 6$, then using  induction, but to no avail. Can someone give me an ideal?
 A: Denote $X = \{x_1,x_2,x_3,...\}=\{1,\;\;2,4,8,16, \;\; 22,24,28,36, \;\; 42,44, ...\}$.
If $x_k \in X$, $k\ge 2$, then $x_k+20 \in X$.
Then
$$4^{odd} \in \{4+20m, \;m \in\mathbb{N}_0 \}\subset X,$$
$$4^{even} \in \{16+20m, \; m\in \mathbb{N}_0\} \subset X.$$
Simply, $4^{odd} \equiv 4 (\bmod 20)$, $4^{even} \equiv 16 (\bmod 20)$.

How one can prove that $x_k\in X \Rightarrow x_k+20 \in X$.
Indeed, for $k\ge 2$: $x_{k+4}=x_k+20.$
Explanation: if the last digit o f$x_k$ is $2$, then
$x_{k+1} =x_k+2$ (with last digit $4$), 
$x_{k+2} = x_{k+1}+4=x_k+6$ (with last digit $8$),
$x_{k+3} = x_{k+2}+8=x_k+14$ (with last digit $6$),
$x_{k+4} =x_{k+3}+6=x_k+20$ (with last digit, of course, $2$).
Similar way, if last digit of $x_k$ is $4$, $8$ or $6$. (note that this sequence contain no $x_k$ with last digit $0$).

On $4^{odd}\equiv 4 (\bmod 20)$, $4^{even}\equiv 16 (\bmod 20)$:
$4^1\equiv 4 (\bmod 20)$, $4^2 \equiv 16 (\bmod 20)$.
Using this base (for math.induction), we will suppose (for $j\in\mathbb{N}$) that 
$4^{2j-1}\equiv 4(\bmod 20)$ and $4^{2j}\equiv 16(\bmod 20)$.

And prove, that 
$4^{2j+1}\equiv 4 (\bmod 20)$, $4^{2j+2}\equiv 16 (\bmod 20)$. 

Proof is almost obvious: $4^{2j+1}\equiv 4^{2j}\times 4\equiv 64\equiv 4 (\bmod 20)$, 
$4^{2j+2}\equiv 4^{2j+1}\times 4 \equiv 16 (\bmod 20)$.
