Question

(a) Show that${10}^n\equiv 1\left(mod9\right)$ for every positive $n$.

(b) Prove that every positive integer is congruent to the sum of its digits mod 9 [For example, $38\equiv 11\left(mod9\right)$.

Short answer

1. It is proved that${10}^n\equiv 1\left(mod9\right)$ for every positive integer$n$ .
2. It is proved that every positive integer is congruent to the sum of its digit’s mod 9.

Step-by-step solution

Prove that 10n≡1 mod 9

We have to show that${10}^n\equiv 1\left(mod9\right)$ .We can provethis by showing for every integer $\overline{)9}\left({10}^n-1\right)$.

Factorize the expression${10}^n-1$ asfollows:

$\begin{array}{rcl}{10}^n-{\left(-1\right)}^n& =& \left(10-1\right)\left(1+10+{10}^2+.....+{10}^{n-1}\right)\\ & =& 9\left(1+10+{10}^2+.......+{10}^{n-1}\right)\end{array}$

According to the divisibility definition, we can say that$\overline{)9}\left({10}^n-1\right)$ .

By congruency definition, we have${10}^n\equiv 1\left(mod9\right)$ .

Hence,it is proved that${10}^n\equiv 1\left(mod9\right)$ for every positive integer$n$ .

Prove that every positive integer is congruent to the sum of its digit’s mod 9 

Let us consider that $n=a_n{a}_{n-1}....a_1{a}_0$, where $a_{1}s$are digits. This implies that we can write n as follows:

$\begin{array}{rcl}n& =& {10}^n{a}_n+{10}^{n-1}{a}_{n-1}+.....+100a_2+100a_1+a_0\\ & =& \left(99...9+1\right)a_n+....+\left(99+1\right)a_2+\left(9+1\right)a_1+a_0\\ & =& 99...9a_n+...99a_2+9a_1+\left(a_n+.....+a_2+a_1+a_0\right)\\ & =& \left(a_n+....+a_2+a_1+a_0\right)mod9\end{array}$

Hence, it is proved that every positive integer is congruent to the sum of its digit’s mod 9.