Proof that $3 \mid 10^{n+2} - 2*10^n + 7, \forall n \in \mathbb{Z}^+$.

Question

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.

Answer 1

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*}

Answer 2

Not by induction, but notice that $10\equiv 1\pmod 3$.

\begin{align}10^{n+2}-2\cdot10^n+7&\equiv1\pmod 3-2\pmod 3+1\pmod 3\\&\equiv 0\pmod 3\end{align}

and we are done.

Answer 3

You almost have it. Starting from your second last line, to try to get the equation into the similar form of $P(k)$, we can do the following,

$$\begin{equation}\begin{aligned} 10^{k+2}(10) - 2\times 10^{k}(10) + 7 & = 10^{k+2}(10) - 2\times 10^{k}(10) + 7(10) - 63 \\ & = 10(10^{k+2} - 2\times 10^{k} + 7) - 7(3^2) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since we've assumed that $P(k)$ is true, i.e., $3 \mid 10^{k+2} - 2\times 10^{k} + 7$, then \eqref{eq1A} shows that $3 \mid 10^{k+3} - 2\times 10^{k+1} + 7$, i.e., $P(k+1)$ is also true.

Answer 4

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.