<s>[INST] Proof that $3 \mid 10^{n+2} - 2*10^n + 7, \forall n \in \mathbb{Z}^+$. This is what I have so far.
Proof by Induction. Let $n \in \mathbb{Z}^+$ Let $P(n)$ be the statement that $10^{n+2} - 2*10^n + 7$ is divisible by 3.
($\textit{Base Case}$): Let $n = 1$.
$$ 
10^{1+2} - 2*10^1 + 7 = 1000 - 20 + 7 = 987
$$
$3 \mid 987$ there for $P(1)$ is true.
($\textit{Inductive Step}$): Let $k \in \mathbb{Z}^+$. Suppose $P(k)$ is true. Now we must show that $P(k+1)$ is true.
$$
10^{(k+1) + 2} - 2*10^{k+1} + 7
$$
$$
\Rightarrow 10^{(k+2)+1} - 2*10^{k+1} + 7
$$
$$
\Rightarrow 10^{k+2}(10)  - 2*10^{k}(10) - 7
$$
$$
\Rightarrow 10(10^{k+2} - 2*10^{k}) + 7
$$
I don't know how to proceed. I've tried other methods of manipulating the equation and nothing seems to work.
 [/INST] With the inductive step note that we have:
\begin{align*}
& 10^{k+2} -2\cdot10^{k} + 7 = 3m\quad\text{for some }m\in \mathbb{Z}\\ 
\implies & 10^{k+2} -2\cdot10^{k} = 3m-7 \qquad(*) 
\end{align*}
Now for the case P(k+1):
\begin{align*}
10^{(k+1)+2} -2\cdot10^{k+1} + 7 & = 10\bigg( 10^{k+2} -2\cdot 10^{k}\bigg) + 7\\
  & = 10\bigg(3m-7\bigg) + 7 \qquad \text{using }(*)\\
  & = 3(10m-21).
\end{align*}
</s>