(10^x - 1) / 9 = y, x member of natural numbers implies y is a natural number.

Proof.

When x = 1 then y = (10^1 - 1) / 9 = (10 - 1) / 9 = 9 / 9 = 1

Assume that (10^a -1) / 9 = b for some a, b members of the natural numbers.

Now consider

c = (10^(a+1) - 1) / 9

c = (10^a.10 - 10 + 9) / 9

c = (10.(10^a - 1) + 9 ) / 9

c =10.(10^a - 1)/9 + 9/9

c = 10.b + 1

Since b is a natural number then so is c.

So, by induction, we have shown that (10^x - 1) / 9 is a natural number or, equivalently, 10^x - 1 is always divisible by 9.

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