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How would I go about proving the following conjecture:

Let $a$ and $b$ be positive natural numbers such that $b>a$ and $\gcd(a,b)=1$. Then $$\{ak\bmod{b}:k\in\mathbb{Z},1\leq k\leq b\}=\{0,1,2,\ldots,b-1\}.$$

I've computed a number of examples. For example, let $a=2$ and $b=5$. Then the set of multiples of $a$ mod $5$ are $2,4,1,3,0$ in order.

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    No...take $a=2,b=17$ just for one example. Or $a=2,b=31$. – 2017-02-10
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    I think you are asking about [Primitive Roots](https://en.wikipedia.org/wiki/Primitive_root_modulo_n) – 2017-02-10
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    That conjecture is false. – 2017-02-10
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    ...and you mean remainders modulo $\;b\;$, right? – 2017-02-10
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    I don’t think that OP is asking about primitive roots; rather I think that (s)he is asking whether $\{na\}$ exhausts the residue classes modulo $b$, under the condition that $\gcd(a,b)=1$. Surely this is correct. – 2017-02-10
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    @Lubin Oh! You may well be right. In that case, of course it is correct. The euclidean algorithm, in the shape of [Bezout's identity](https://en.wikipedia.org/wiki/B%C3%A9zout's_identity) shows this. – 2017-02-10
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    oh cool.thanks a lot. will look into it. – 2017-02-10
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    yeah lubin you are right. apologies for the unclearity. – 2017-02-10
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    @user367810 I edited your post to make it more clear what you were asking, since a number of people appeared to misread it. – 2017-02-10
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    thanks a lot Stella Biderman – 2017-02-10
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    @user367810 if any of the answers satisfy you, you should accept one of them by clicking the checkmark. You can also upvote any you like by pressing the up arrow, but you can only accept one answer. – 2017-02-10

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Here's a simple proof: Assume $am=an\pmod{b}$ with $0\leq m

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Suppose that it isn't true. Then there exists $k$ and $j$ that are integers such that $1\leq k