Question

Prove that${10}^n\equiv {\left(-1\right)}^n\left(mod11\right)$ for every positive$n$ .

Short answer

It is proved that${10}^n\equiv {\left(-1\right)}^n\left(modn\right)$ for every positive integer$n$ .

Step-by-step solution

Prove that 10n≡-1n mod 11 

We have to show that ${10}^n\equiv {\left(-1\right)}^n\left(modn\right)$.We can prove this by showing for every integer $\overline{)11}\left({10}^n-{\left(-1\right)}^n\right)$.

Factorize the expression ${10}^n-{\left(-1\right)}^n$as follows:

$\begin{array}{rcl}{10}^n-{\left(-1\right)}^n& =& \left(10-\left(-1\right)\right)\left({10}^{n-1}+{10}^{n-2}\left(-1\right)+{10}^{n-3}{\left(-1\right)}^2+....+{\left(-1\right)}^{n-1}\right)\\ & =& \left(10+1\right)\left({10}^{n-1}+{10}^{n-2}\left(-1\right)+{10}^{n-3}{\left(-1\right)}^2+....+{\left(-1\right)}^{n-1}\right)\\ & =& 11\left({10}^{n-1}+{10}^{n-2}\left(-1\right)+{10}^{n-3}{\left(-1\right)}^2+.....+{\left(-1\right)}^{n-1}\right)\end{array}$

Apply congruency definition

According tothedivisibility definition, we can say that$\overline{)11}\left({10}^n-{\left(-1\right)}^n\right)$ .

By congruency definition, we have${10}^n\equiv {\left(-1\right)}^n\left(modn\right)$ .

Hence, it is proved that ${10}^n\equiv {\left(-1\right)}^n\left(modn\right)$for every positive integer $n$.