Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots 
Let $\overline{a_n \ldots a_1a_0}$ be the decimal representation of $65^k$ for some $k \geq 2$. Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots.

I thought about using the rational root theorem. We know that $a_0 = 5$ since $65$ ends in a $5$, so any rational root must be of the form $-\dfrac{5}{a}$ or $-\dfrac{1}{a}$ where $a$ is an integer factor of $a_n$. How can we continue?

Let $f_n$ be the polynomial obtained from the digits of $65^n$. The other answer shows that if $p/q$ is a root of this polynomial in lowest terms, then $p-10q|65^n$; along with the bounds on $f_n$, this implies that we must have $p=-5$ and either $q=2$ or $q=6$. That is, the only possible rational roots are $-\frac{5}{6}$ and $-\frac{5}{2}$.
Note that $(6x+5)^n$ is also a polynomial which evaluates to $65^n$ at $x=10$. So we can write
$$
f_n(x)=(6x+5)^n+(x-10)R(x)
$$
for some polynomial $R$.
Setting $x=0$ in this expression, we have
$
5=5^n-10R(0)
$,
and so $R(0)=\frac{5^{n-1}-1}{2}$, which is not a multiple of $5$.
Now, suppose $n \geq 3$ and $x$ is either $-\frac{5}{6}$ or $-\frac{5}{2}$. Then $(6x+5)^n$ has $5$-adic valuation at least $3$ (that is, its numerator is a multiple of $5^3$ when it is written in lowest terms). In contrast, the $5$-adic valuation of $R(x)$ is $0$ (because the $5$-adic valuation of $R(0)$ is $0$) and $(x-10)$ is either $-\frac{5}{6}-10=-\frac{65}{6}$ or $-\frac{5}{2}-10=-\frac{25}{2}$, neither of which has $5$-adic valuation greater than $2$. So $(6x+5)^n$ and $(x-10)R(x)$ have unequal $5$-adic valuations, which means that $f_n(x)=(6x+5)^n+(x-10)R(x)$ cannot possibly be $0$.
It only remains to check that $f_2$ has no rational roots, which is an easy calculation.
