Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$.

Question

$$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$

My proof of $(\star)$ is as follows:

\begin{align} 10^{2n+1}+1 &= 10\cdot10^{2n}+1 \\ &= (11-1)\cdot10^{2n}+1 \\ &= 11\cdot10^{2n}-10^{2n}+1 \\ &= 11\cdot10^{2n}-\left(10^{2n}-1\right) \\ &= 11\cdot10^{2n}-\left(100^{n}-1\right) \\ &= 11\cdot10^{2n}-\left((99+1)^{n}-1\right) \\ &= 11\cdot10^{2n}-\left(1+\binom{n}{1}99+\binom{n}{2}99^2+\cdots+\binom{n}{n-1}99^{n-1}+99^n-1\right) \\ &= 11\cdot10^{2n}-\underbrace{99}_{11\cdot9} \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right) \\ &= 11\left(10^{2n}-9 \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right)\right) \end{align}

Is there an easier way to prove $(\star)$? The expansion of $(99+1)^n$ seems unnecessarily complicated, but I wasn't sure how else to go from there. Easier proofs are welcome!

Answer 1

Use modular arithmetic:

$$10\equiv-1\mod11$$

$$10^2\equiv1\mod11$$

$$10^{2n}\equiv1\mod11$$

$$10^{2n+1}\equiv-1\mod11$$

$$11|10^{2n+1}+1$$

Answer 2

Because for $n\geq1$ we have: $$10^{2n+1}+1=(10+1)(10^{2n}-10^{2n-1}+...-10+1).$$ For $n=0$ it's obvious.

Answer 3

$2n+1$ is odd and $10\cong -1\pmod {11}$. Thus $10^{2n+1}\cong (-1)^{2n+1}\cong -1\pmod{11}$.

Answer 4

Not as easy as the other answers, but this is a classic induction question.

For $n = 0$, we have $10^{2n + 1} + 1 = 11$, which $11$ divides, confirming the base case.

Suppose $11$ divides $10^{2n + 1} + 1$ for some $n$. Then some $k$ exists such that $10^{2n + 1} + 1 = 11k$. Then, $$10^{2(n + 1) + 1} + 1 = 100 \cdot 10^{2n + 1} + 1 = 100 (11k - 1) + 1 = 11(100k - 9),$$ thus $11$ divides $10^{2(n+1) + 1} + 1$. By induction, $11$ divides $10^{2n + 1} + 1$ for all $n$.

Answer 5

Use the identity $a^{2n+1}+b^{2n+1} =(a+b)\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j} $.

Then put $a=10, b=1$.

Proof.

$\begin{array}\\ (a+b)\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j} &=a\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j}+b\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j}\\ &=\sum_{j=0}^{2n} (-1)^j a^{j+1} b^{2n-j}+\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j+1}\\ &=\sum_{j=1}^{2n+1} (-1)^{j-1} a^{j} b^{2n-j+1}+\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j+1}\\ &=\sum_{j=1}^{2n} (-1)^{j-1} a^{j} b^{2n-j+1}+(-1)^{2n} a^{2n+1} b^{0}+\sum_{j=1}^{2n} (-1)^j a^j b^{2n-j+1}+(-1)^0 a^0 b^{2n+1}\\ &=\sum_{j=1}^{2n} ((-1)^{j-1}+(-1)^j) a^{j} b^{2n-j+1}+a^{2n+1}+ b^{2n+1}\\ &=a^{2n+1}+ b^{2n+1}\\ \end{array} $

Answer 6

$$10^{2n+1}+1 \equiv (-1)^{2n+1} + 1 \equiv -1 + 1 \equiv 0 \pmod{11}.$$