Japan competition problems, 2015

For integers $k$, $\ell$ and a positive integer $m$, let us write $k \equiv \ell \pmod m$ to mean that $k - \ell$ is divisible by $m$. For a non-negative integer $n$, let $a_n$ be the integer obtained by applying the operation stated for the problem $n$ times starting with the initial number $a_0 = 25^{2015}$. Let for a non-negative integer $n$ $b_n$ be the lowest digit number of $a_n$. Then, we get
$$a_{n+1} = \frac{a_n - b_n}{10} + 4b_n = \frac{a_n + 39b_n}{10}$$
Therefore, we get $a_{n+1} \le \frac{a_n}{10} + \frac{39}{10} \cdot 9$, from which it follows that $a_{n+1} - 39 \le \frac{a_n - 39}{10}$.
It follows further that $a_n - 39 \le \frac{a_0 - 39}{10^n}$ is valid for any $n$. We also have
$$a_{n+1} = 4a_n - 39 \cdot \frac{a_n - b_n}{10} \equiv 4a_n \pmod{39}$$
Also from $a_0 = 25^{2015} < 10^{2015} = 10^{4030}$, we get $a_{10000} - 39 < \frac{a_0 - 39}{10^{10000}} < 1$, which implies that $a_{10000} \le 39$. From $a_{10000} = 4^{10000}a_0 = 4^{10000} \cdot 25^{2015} \pmod{39}$, we

$$a_{10000} \equiv 1 \pmod{3}$$
$$a_{10000} \equiv 4 \cdot 64^{3333} \cdot (-1)^{2015} \equiv 4 \cdot (-1)^{3333} \cdot (-1)^{2015} \equiv 4 \pmod{13},$$
which says that $a_{10000}$ is a positive integer less than or equal to 39 and has the remainder 1 when divided by 3 and remainder 4 when divided by 13. It is easy to check that 4 is the only such integer.