Proving (10^m)-1 is divisible by 9

Wow... didn't think of that, I used the google calculator and now I feel like a dumbass -.-. Thanks guys, feel better now. Here's another one I'm having a little difficulty with, and I don't feel like spamming these forums.

Basically the part I'm having difficulty with is a portion of a larger proof. In this section of the aforementioned proof I have to show that (10 ^ m) - 1, for any given integer m, will be divisible by 9. I can easily solve it from example, but I want to be able to show that it will be true for any arbitrarily chosen m.. I guess it's a question of phrasing. Using the definition of divisibility I get..

9 k = (10 ^ m) - 1 , and I should show that k must be an integer. Some help with this one would be nice
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 Quote by Animuo Wow... f'ing calculator made me feel like a dumbass -.-. Thanks guys, feel better now. Here's another one I'm having a little difficulty with, and I don't feel like spamming these forums. Basically the part I'm having difficulty with is a portion of a larger proof. In this section of the aforementioned proof I have to show that (10 ^ m) - 1, for any given integer m, will be divisible by 9. I can easily solve it from example, but I want to be able to show that it will be true for any arbitrarily chosen m.. I guess it's a question of phrasing. Using the definition of divisibility I get.. 9 k = (10 ^ m) - 1 , and I should show that k must be an integer. Some help with this one would be nice

The easiest way to do this is to use modular arithmetic. Are you familiar with that? Start with $10 \equiv 1 \pmod 9$.

If you don't know how to work with that, the elementary way of showing this is to simply consider what happens when you subtract one from a number with 1 followed by nothing but zeroes. Then use the divisibility rule for 9.

A final way to do it is to use Remainder Theorem to figure out the remainder when the polynomial $x^m - 1$ is divided by $x-1$, then substitute a suitable value for x to prove your case.
 Yeah I'm just getting to it now actually, but problem I mentioned is prior to the discussion of mod and div in the book. I suppose it expects me to just spell out what happens when one is subtracted from any number 10 ^ m could produce, but I thought there would be a general more axiomatic truth that I could use.. I guess that's what mod is, I'll read on, thanks for the heads up.

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Proving (10^m)-1 is divisible by 9

 Quote by Animuo Yeah I'm just getting to it now actually, but problem I mentioned is prior to the discussion of mod and div in the book. I suppose it expects me to just spell out what happens when one is subtracted from any number 10 ^ m could produce, but I thought there would be a general more axiomatic truth that I could use.. I guess that's what mod is, I'll read on, thanks for the heads up.
I hope you took note of my final edit - one more way to do it that you might be more familiar with.
 Recognitions: Homework Help One more way is to remember that, for example, $999 = 9 \times 10^2 + 9 \times 10^1 + 9 \times 10^0= 9(10^2 + \times 10^1 \times 10^0)$, so if you can get $10^m - 1$ in this form, you will have shown that it is divisible by 9. To that, just use $10^m-1 = 10^m -10 + 9 = 10(10^{m-1}-1) + 9 \times 10^0$ iteratively until the exponent becomes m-m=0.

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 Quote by Animuo I have to show that (10 ^ m) - 1, for any given integer m, will be divisible by 9.
Well it's not true for just any integer m, m=-1 gives us a fraction

So we want to prove this is true for all integers $m\geq 0$. Induction is a relatively simple way of proving this.
 Recognitions: Gold Member Consider using the binomial theorem with 10m = (1+9)m
 Hint for mathematical induction: $$10^{m + 1} = 10 \, 10^{m} = 10 (9 k_m + 1) = 90 k_m + 10 = 9 (10 k_m + 1) + 1$$
 I'll bet Binomial theorem is useful for proving divisibility with n if the desired form can be expressed in the form (nk ± 1), just expand and manipulate to get answer.