Proving a positive integer Suppose that a positive integer $n$ ends with $k$ $9\text{'s}$. How can we prove that the integer $3n^4+4n^3$ ends with at least $2k$ $9\text{'s}$?

Firstly note that saying a number ends with $k$ $9$s is equivalent to saying that it is congruent $-1\ (\text{mod }10^k)$, as for example $99$ ends with two $9$s and $99\equiv -1\ (\text{mod }100)$.
So write $n=A\cdot 10^k -1$, for an integer $A$. Then note:
\begin{align}
n^3&\equiv 3A\cdot10^k-1\ \quad&(\text{mod }10^{2k})\\
n^4&\equiv -4A\cdot10^k+1 \quad&(\text{mod }10^{2k})
\end{align}
As all the other terms include $10$ to a $\geq2k$ integer power. So adding:
\begin{align}
3n^4+4n^3 &\equiv (-12A\cdot10^k+3) + (12A\cdot10^k-4)\quad&(\text{mod }10^{2k})\\
&\equiv -1 \quad&(\text{mod }10^{2k})
\end{align}
Which is the required answer.
