202009290227 Exercise 2.3.5

Prove that $10^{n+1}+10^n+1$ is divisible by $3$ for $n\in\mathbb{N}$ .

_{Extracted from T. W. Judson. (2021). Abstract Algebra Theory and Applications.}

Proof.

Let $P(n)$ be the statement:

$P(n):\qquad 3\,\Big| (10^{n+1}+10^n+1)$ for any $n\in\mathbb{N}$

Determine whether or not $P(n)$ is true when $n=1$ :

$P(1): \qquad 3\,\Big| (10^{(1)+1}+10^{(1)}+1)$

As $10^{(1)+1}+10^{(1)}+1 = 111 = 37\cdot 3$ is divisible by $3$ , $P(1)$ is true.

Suppose $P(n)$ is true for some $n\, (\geqslant 1) \in\mathbb{N}$ , try and prove the statement $P(n+1)$ :

$P(n+1): \qquad 3\,\Big| (10^{(n+1)+1}+10^{(n+1)}+1)$

$\begin{aligned} &\quad 10^{(n+1)+1}+10^{(n+1)}+1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +10 - 9 \\ = &\quad 10\cdot (10^{n+1}+10^n+1) - 9 \\ \end{aligned}$

As $P(n)$ is true and by the fact that three divides nine, $10\cdot (10^{n+1}+10^n+1) - 9$ is therefore divisible by $3$ . That is,

$P(n)\textrm{\enspace is true\enspace}\Rightarrow P(n+1)\textrm{\enspace is true\enspace}$

That $P(n)$ is true for $n=1$ , by the principle of mathematical induction, I have thus proven $10^{n+1}+10^n+1$ is divisible by $3$ for $n\in\mathbb{N}$ .

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