TITLE: Starting with $13^{2013}$ can we get $2013^{13}$ by the following process.
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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.