Need some help regarding the family of $\mathbb{Z}_n = \lbrace 0, 1, 2, 3, \ldots, n-1 \rbrace$

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I have a couple of questions about some problems in my discrete mathematics text Discrete Mathmatical Structures 6th Edition, on page 467-468 they are as follows.

Each member of the family $\mathbb{Z}_n$ has two operations defined as follows:

$a \oplus _n b = a + b (mod n)$ & $a \otimes_n b = ab (mod n)$

The result of each operation mod $n$ must be a number between $0$ and $n - 1$ (inclusive), so to satisfy the closure property for each operation we restrict the objects in the structure based on mod n to $0, 1, 2, ..., n - 1$. Let $\mathbb{Z}_n = (\lbrace0, 1, 2, 3, ...., n-1\rbrace, \oplus_n, \otimes_n).$

Part 3: Here you will develop some general conclusions about the family of $\mathbb{Z}_n$.

1: Let $a ∈ \mathbb{Z}_n$ and $a \neq 0$. Tell how to compute -a using n and a.

2: For which positive integers k does $a \otimes_k x = 1$ have a unique solution for each a, 0 < a < k -1?

3: For which positive integers k does $a \otimes_k x = 1$ not have a unique solution for each a, 0 < a < k -1?

For the first question I had no idea how to proceed. However, a friend asked a professor who said the answer was $-a = an - a[n+1]$, I would like to understand the process of how they came to that answer.

For the second two answers I am confused on what the text is asking for and what I need to do to find the answer. These operations were never covered in the text. I've tried just picking an $a$ and a $k$ and then proceed to solve for $x$ but I get stuck because I don't know how to remove a $(mod)$ function algebraically.

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For part 1, notice that what you need to do is essentially break $\mathbb{Z}$ into sets of length $n$, starting from 0, and going in both directions. Then, notice that for some $a$, you can get a nice relation for $-a$ as well, think of it as finding the position of $-a$ in this set of $n$ elements. You might also just use the addition operation and use the fact that $-a$ is the additive inverse of $a$.

For part 2, I am not very sure about what you mean. Suppose that $k|(ax-1)$. If $x_0$ is a solution to this, note that $x_i=x_0+ik, i\in \mathbb{Z}$ also satisfies your equation. I believe there must be some restrictions on $x$ as well?