Prove that for every integer n ≥ 0 ,   10 n = 1 ( mod   9 ) .

1. Use part (a) to prove that a positive integer is divisible by 9 if, and only If, the sum of its igits is divisible by 9.

(a)

To prove:

The statement 10n≡1(mod9) for all integers n≥0 ,

Given information:

The given statement is 10n≡1(mod9).

Formula used:

Division algorithm:

Let a be an integer and d be a positive integer. Then, there are unique integers q and r with 0≤r<d such that a=dq+r, here, q is the quotient and r is the remainder.

So, q=a div d and r=amodd.

When m divides a−b then a is congruent to b modulo m. The notation is a≡b(modm).

The base b representation of n is ak…a2a1a0, then n=akbk+ak−1bk−1+…+a1b+a0.

a divides b if there exists an integer c such that b=ac. The notation is a|b.

Proof:

To prove the statement 10n≡1(mod9), by using mathematical induction on n.

Let P(n) be 10n≡1(mod9).

For n=0 ,

By the definition, 9 divides 100−1.

100−1=1−1=0

9|(100−1)

Here, 0 is an integer

(b)

To prove:

A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9.