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How can I show that $10^n-1, 10^{n-1}-1,...., 10-1$ are all divisible by 9? I was considering using Euclid's algorithm, but I can't find a way to get that to work.

  • Hint: use induction, $10^{n+1}-1=10\cdot 10^n-1=10\cdot (9k+1)-1=\ldots$. – Git Gud Sep 18 '13 at 06:26
  • I would use induction – Spaderdabomb Sep 18 '13 at 06:27
  • I don't get the third part, where did the power go? –  Sep 18 '13 at 06:35
  • Your induction hypothesis is that $10^n-1$ is divisble by $9$, that means that there exists an integer $k$ such that $10^n-1=9k$, thus $10^n=9k+1$. Do you understand this? – Git Gud Sep 18 '13 at 06:40
  • Yeah I got it, thank you. –  Sep 18 '13 at 06:49
  • http://math.stackexchange.com/questions/495597/divisibility-for-natural-numbers – lab bhattacharjee Sep 18 '13 at 07:57
  • I dont find induction very satisfying as far as this particular problem goes. What about this? 10^n is a 1 followed by n zeroes. 10^n-1 is a string of 9s. Such a number can also be written as a string of 1s multiplied by 9. Do write what you think. – Adam Nov 18 '13 at 13:00

4 Answers4

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$$ a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) $$

Proof of the identity by expansion:

$$\begin{align} &(a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \\\\ &=a^{n} + a^{n-1}b + a^{n-2}b^2 + \ldots + ab^{n-1}\\\\ &\;\;\;\;\;\;\;\;- ba^{n-1} - b^2a^{n-2}-\ldots -b^{n-1}a - b^n\\\\ &=a^n-b^n \end{align}$$

Parth Thakkar
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    Not that this is wrong, but in this context one would require a proof of the identity above and that goes by induction. However, if you're gonna have to use induction, it's easier to just prove the particular case asked in the question. – Git Gud Sep 18 '13 at 06:28
  • @GitGud, to prove the identity, all we need to do is to expand. Why use induction for that? – Parth Thakkar Sep 18 '13 at 06:29
  • Expand what? The RHS? You can't do computations with $\ldots$ in the middle. Everything that has $\ldots$ in it requires induction. – Git Gud Sep 18 '13 at 06:31
  • @GitGud: Ostensibly, this identity doesn't require proof because one has already learned it in their algebra courses. –  Sep 18 '13 at 06:34
  • @GitGud, is that fine now? – Parth Thakkar Sep 18 '13 at 06:37
  • @Hurkyl It's possible, but I don't think one would be dealing with a problem such as the one at hand after attenting an algebra course. – Git Gud Sep 18 '13 at 06:37
  • @GitGud, that, I have agree upon with you :D – Parth Thakkar Sep 18 '13 at 06:37
  • @ParthThakkar No, I stand by my statement: you can't do computations with $\ldots$ in the middle. Everything that has $\ldots $ in it requires induction. – Git Gud Sep 18 '13 at 06:38
  • @GitGud: By "algebra" I meant the high school course, not abstract algebra. –  Sep 18 '13 at 06:39
  • @Hurkyl Then I forward you to my statement above. – Git Gud Sep 18 '13 at 06:41
  • @Hurkyl, such a thing won't be dealt in abstract algebra I guess. I know you are defending my answer and I must remain shut over here, but couldn't help pointing out. And Gitgud, I think such a thing would be definitely taught in High Schools. (I learnt it last year or maybe the year before). – Parth Thakkar Sep 18 '13 at 06:43
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    @GitGud Actually, this can be proven with a simple summation and an indexed variable. It is usually best to shy away from induction when at all possible because it has the tendency to prove conditions which are sufficient, but not those which are necessary. – Ryan Sep 18 '13 at 06:45
  • @Ryan Oh really? Please show me a proof. I'll read it later. I'm off now. – Git Gud Sep 18 '13 at 06:50
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    @GitGud $$\begin{align}(a-b)(a^{n-1}+a^{n-2}b+\cdots +b^{n-1}) &= (a-b)\left(\sum_{i=0}^{n-1}a^{n-1-i}b^i\right)\ &= \sum_{i=0}^{n-1}a^{n-i}b^i-\sum_{i=0}^{n-1}a^{n-1-i}b^{i+1}\ &= \sum_{i=1}^n a^ib^{n-i}-\sum_{i=0}^{n-1} a^ib^{n-i}\ &= a^n+\sum_{i=1}^{n-1} a^ib^{n-i}-\sum_{i=1}^{n-1} a^ib^{n-i}-b^n\ &= a^n-b^n.\end{align}$$ Where we have only use standard properties of $\sum$ given that multiplication is distributive towards adition and that sums are commutative. – Josué Tonelli-Cueto Sep 18 '13 at 14:47
  • @JosuéTonelli-Cueto How do you prove distributivity,for instance, without induction? This is just delaying the issue. Induction is unavoidable. – Git Gud Sep 18 '13 at 14:52
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    Is avoidable, since I have not used it directly. That's like saying that foundations of set theory are unavoidable in order to prove that since without them induction cannot be proven. – Josué Tonelli-Cueto Sep 18 '13 at 14:59
  • Of course induction is impossible to avoid, it lies on the foundations. But in a certain sense, you're making a very imediate use of induction. You used it, you just didn't say you did.You assumed distributivy. Might as well assume the identity itself. – Git Gud Sep 18 '13 at 15:34
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    @Git: Of course induction is not impossible to avoid: all you need is some simple axioms about summations, and Josue shows how to make the proof. You can, of course, add induction to anything you want (you used plus? you must have inducted over the successor function!), but that doesn't mean induction was there in the original argument. We build mathematical tools so that we can solve problems with them -- not so that we can forget about the tools and solve problems with the building blocks instead. –  Sep 18 '13 at 20:04
  • @Hurkyl Changing the standard foundations just to make it look like you're right. OK... – Git Gud Sep 20 '13 at 23:04
  • @Git: Who's changing foundations? We're manipulating summations according to familiar algebraic identities; this is no different than invoking, say, $a+b = b+a$ in an ordinary algebraic calculation. Unless, of course, you're in the peculiar (and brief!) situation of being in the business of studying foundations and are at the point where you have already proven commutativity of $+$ but have not yet proven the usual summation identities. –  Sep 21 '13 at 11:09
  • @Hurkyl I don't think you will want to take this discussion as deep as I wish, therefore I'm not replying back regarding this. – Git Gud Sep 21 '13 at 11:42
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    @JosuéTonelli-Cueto: I've edited an above comment of yours so that it doesn't break the page layout. No content should have changed, but double check to be certain. – user642796 Sep 21 '13 at 13:59
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we can start by 10 ≡ 1 (mod 9)

and $10^n$ ≡ 1 (mod 9) for all {n≥1}

follows from the property of congruence

If a≡ b(mod n)then $a^r≡ b^r$ (mod n), for any integer r≥ 1

$10^n-1$ ≡ 0 (mod 9)

implying that $10^n-1$ is divisible by 9 for all {n ≥ 1}

so you proved it

Mr. Math
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  • This is circular. – Git Gud Sep 18 '13 at 06:27
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    what do u mean circular this just follows the properties of modular arithmetic If a≡ b(mod n)then $a^r≡ b^r$ (mod n), for any integer r≥ 1 – Mr. Math Sep 18 '13 at 06:31
  • Circular in the sense that to use those properties, you'll need to first prove some other properties which allow you to conclude what's asked immediately. Edit: and if it isn't circular, it is at least a huge overkill and not what's intended with this problem. – Git Gud Sep 18 '13 at 06:32
  • you dont need to only condition is n≥ 1 If you look at any number theory book it tells you that this is an independent property – Mr. Math Sep 18 '13 at 06:34
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    @GitGud: The ability to solve problems like this with ease are one of the main reasons to learn about modular arithmetic.... –  Sep 18 '13 at 06:36
  • @Hurkyl Of course. If you ignore pedagogy this is probably fine. But there's no way this problem would come up after modular arithmetic. – Git Gud Sep 18 '13 at 06:42
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    @GitGud: Then it should. I see plenty of people who have learned modular arithmetic that simply have no concept of reversing the definition -- they continue to do laborious integer algebraic calculations to solve divisibility problems, they convert modular arithmetic problems into divisibility problems so that they can do laborious integer algebraic calculations, but they have no concept of going the other way: that "Divisibility problem? Maybe I should us modular arithmetic to solve it...." –  Sep 18 '13 at 06:56
  • I think this is relevant. Lets have the problem of sin x/x as x goes to 0. Now someone might say lets use lHosiptal theorem. And someone else might say that to come with the derivative of sine you need to solve the limit. Still another person might come along and say that, while the second person is right, in principle, in practice can save a lot of work doing lHospital even though it might be circular. – Adam Nov 18 '13 at 13:21
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$10^n - 1$ is the number consisting of $n$ nines. Just do long division.

Arthur
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$10^n - 1 = 999\ldots 9 = 9 \cdot 111\ldots 1$

Daniel McLaury
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