Starting with $13^{2013}$ can we get $2013^{13}$ by the following process. This is the problem that I found in a question paper. The problem is:
A positive integer is written on the board. We repeatedly erase its unit digit and add 4 times that digit to what remains. Starting with the number $13^{2013}$ can we ever end up at $2013^{13}$.
I really don't know how to solve these kind of problems.
Firstly I thought that I should start by finding the last digit of $13^{2013}$ but this method is not ideal for such kind of problems.
What should be my approach to solve these kind of problems.
Please help me in learning a new thing in mathematics.
 A: The other option - 
Note that $3 \mid 2013$, and $3 \nmid 13^k$
$10a+b = 9a +(a+b) \equiv (a+b) \bmod 3 \\
a+4b = (a+b)+3b \equiv (a+b) \bmod 3 \\
\therefore 10a+b \equiv a+4b \bmod 3$
The digit-erasure process transforms $10a+b$ into $a+4b$ and preserves the  $\pmod 3$ congruence class
Since $2013^{13}$ and $13^{2013}$ are in different congruence classes $\pmod 3$, the erasure process cannot transform one into the other.
A: To simplify notation, for $n=10a+b$ with $a\in\mathbb{N}$ and $b\in\{0,1,2,\dots,9\}$ define $P(n)=P(10a+b) = a+4b$.  That is to say, $P(n)$ is precisely the function you describe of deleting the final digit and adding 4 times the deleted digit to the rest of the number.
Note that $13|n \Leftrightarrow 13|P(n) \Leftrightarrow 13|P^k(n)$  (try to prove this if you haven't seen it before)
Since we start with $13^{2013}$, clearly $13|13^{2013}$ so by applying $P$ multiple times we ask if we can ever get to $2013^{13}$
Notice however that $13\nmid 2013^{13}$ since $13$ is prime and $13\nmid 2013$ (by the fundamental theorem of arithmetic and since the prime decomposition of $2013$ is $3\cdot 11\cdot 61$)
It follows then that there does not exist a $k$ such that $P^k(13^{2013})=2013^{13}$ since the L.H.S. is a multiple of $13$ while the R.H.S. is not.
