<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] Note that $10^k=99\dots(k\ times)\dots 9+1$. Now $99\dots(k\ times)\dots 9$ is always divisible by $9$ and therefore it is divisible by $3$. So, dividing $10^k$ by $9$ gives a remainder of $1$.
Therefore, $10^{n+2}\equiv 1 \mod 3$ and $10^{n}\equiv 1\mod 3$. So, $10^{n+2}-2\times 10^{n}\equiv 1-2\times 1\mod 3\equiv -1 \mod3$.
And, this implies that $10^{n+2}-2\times 10^{n}+7\equiv -1+7\mod 3\equiv 6\mod 3\equiv 0\mod 3$.
Hence, 3 divides $10^{n+2}-2\times 10^{n}+7$. Here the notation $a\equiv b\mod m$ means that $m$ divides $a-b$.
If you want to go by the method of mathematical induction then notice that you assumed that $P(k)$ is true so $3$ divides $10^{k+2}-2\times 10^{k}+7$. Let $10^{k+2}-2\times 10^{k}+7=3m$ for some integer $m$.
Now, $10(10^{k+2}-2\times 10^{k})+7=10(10^{k+2}-2\times 10^{k}+7-7)+7=10(10^{k+2}-2\times 10^{k}+7)-70+7=10(10^{k+2}-2\times 10^{k}+7)-63=30m-63$.
Here, the first term is divisible by $3$ and $63$ is also divisible by $3$. So, the whole expression is divisible by 3. This would complete your argument by induction.
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