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If it's infinite, is it countable or uncountable infinite?

I am a newbie to this topic... I don't know what modular arithmetic for polynomials means. Can someone please give me a link where I can learn?

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    Thanks. This kind of helps me understand. Although I don't really know what field means and the notation z/7z is new to me.2012-08-26

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There are $8$ coefficients to be determined. The lead coefficient cannot be $0$. So the number is $(6)(7^7)$.

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    Without loss of generality we can take them to be as you described. In principle they are abstract objects $[0], [1], \dots, [6]$.2012-08-26
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You only have 6 or 7 (why not always 7?) choices for the coefficients. Thus the number is finite, and you should be able to figure it out for yourself....

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    @DavidWallace When I answered the post was only consisting of the first question. And since it looked like a simple, potential homework question, I didn't provide a full answer. I guess in the future I should try to guess potential future edits ;)2012-08-26
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All such polynomials look like: $ a_1 x^0 + a_2 x^1 + \cdots + a_n x^{n-1} + a_{n+1} x^{n}$ where $a_i \in \{ \color{blue}{0}, 1, \ldots, n-1 \}$ for $1 \le i \le n$ and $a_{n+1} \in \{1, 2, \dots, n-1 \}.$

So there are $\underbrace{n \times n \times \cdots \times n}_{n\text{ times}} \times (n-1) = n^n \times (n-1)$ such polynomials.