Even Number Divisibility Problem The question is that for the equation $x = 25 + 5^k$ where $k$ is some random positive integer, can $x$ be divisible by $9$ for any $k$?
My first intuition is that since $25$ = odd and $5^k$=odd then $x$ must be an even number. Rule of divisibility by $9$ states that the sum of digits should be divisible by $9$.
Since $x$ is an even number :
The series :  $x = 18 , 36 , 54 , 72... $
Is there any $k$ value that corresponds to any number in this series? I tried to write a Python script and it seems like there are not any. What is the reason behind that?

You want to solve $25+5^k\equiv_9 0$, this is equivalent to $$5^{k-2}\equiv_9 -1\equiv_9 5^3\\5^{k-5}\equiv_9 1$$
We also have $\text{ord}_9(5)=6$, so $6\mid k-5$. Therefore $$25+5^{5+6l}\equiv_9 0$$ That is, $k=5+6l$, with $l$ any non-negative integer is a general solution.
The first few values:
$$\begin{array} {|r|r|}\hline l & k & 25+5^{5+6l} \\ \hline 0 & 5 & 3150 \\ \hline 1 & 11 & 48828150 \\ \hline 2 & 17 & 762939453150 \\ \hline 3 & 23 & 11920928955078150 \\ \hline 4 & 29 & 186264514923095703150 \\ \hline  \end{array}$$
