Proving divisibility without an inductive proof.

Question

Is there a way to prove that $$10^{n+1} + 10^n + 1$$ is divisible by three without using a proof by induction? We are supposed to use the properties of expressions such as $a$ is congruent to $b \pmod m$.

Answer 1

Just note that

$$10^{n + 1} + 10^n + 1 \equiv 1^{n + 1} + 1^n + 1 \equiv 0 \pmod{3}$$

as desired.

Answer 2

Well, how about a proof using the binomial theorem?

$10^m = (1+9)^m = 1 + \sum_{k=1}^m \binom{m}{k} 9^k$. Since $3 \mid 9$, we see that $ 3 \mid (10^m-1)$ for all $m \ge 1$.

Finally, since $10^{n+1}+10^n +1 = (10^{n+1}-1)+(10^n-1) +(1-1) + 3$, it follows that $3 \mid (10^{n+1}+10^n +1) $.