TITLE: A positive integer $k$ such that the last $2015$ digits of $53^k$ are all $3$
QUESTION [9 upvotes]: Show that there is a positive integer $k$ such that the last $2015$ digits of $53^k$ are all $3$.

We need $3 \cdot 53^k \equiv -1 \pmod{10^{2015}}$. I didn't see a way to prove there exists such an integer $k$ because the modulus is so large. How do we prove there exists such an integer $k$?

REPLY [3 votes]: Once we find a power of $53$ whose last four digits are $3$, we can get the desired $k$. Specifically, let $a$, $k_4$ be integers such that $53^{k_4}\equiv a\mod 10^4$. I claim that for every integer $n\geq 4$, there is a positive integer $k_n$ such that $$53^{k_n}\equiv a\mod 10^n.$$
The proof is induction on $n$. The base case $k_4$ exists by hypothesis. Now, let $n\geq 4$ and suppose $53^{k_{n}}\equiv a\mod 10^n$. We will have $53^{k_n}= a+10^nb$ for some integer $b$. We can compute directly that $53^{500}\equiv 1+10^4\mod 10^5$. It follows that 
$$
53^{500\cdot 10^{n-4}b a^{-1}}\equiv 1+10^n ba^{-1}\mod 10^{n+1},
$$
where $a^{-1}$ denotes any inverse of $a$ modulo $10$. We take $k_{n+1}=k_n-500\cdot 10^{n-4}ba^{-1}$ to get
$$
53^{k_{n+1}}\equiv (a+10^nb)(1-10^n ba^{-1})\equiv a\mod 10^{n+1}.
$$
To finish the problem, check that
$$
53^{29}\equiv 3333\mod 10^4.
$$