I have the following question which I do not understand. Here it is:
Consider the primes $5$, $7$ and $11$ as n. For each integer from $1$ through $n - 1$, calculate its inverse.
I do not understand what this question is exactly saying. Would I only have to do $1$ through $(11-1)$ and find the inverse of those?
Or is it asking me to find the inverse for the function $n-1$?
If someone could clear this up for me that'd be great!
On
I think you are needing to work with modulo arithmetic. The term inverse though can mean a few things. Explanation below.
Additive inverse: $a+(-a)=0$ here we have that $-a$ is the additive inverse of $a$. I don't think you are interested in additive inverses here. Additive inverses mod n are from the group $\mathbb{Z}_n=\{0,1,2,...,n-1\}$.
Multiplicative inverses: $a(b)=1$. Here we have that $a^{-1}=b$, or that the inverse of $a$ is $b$. You are probably interested in multiplicative inverses, since in mod n they only make sense on the group $\mathbb{Z}_n^*=\{1,2,3,...,n-1\}$. This is because $0$ does not have a multiplicative inverse. That is there is not an $x$ such that $0(x)=1$.
With small groups, like $\mathbb{Z}_{11}$, you can find the inverses by hand by brute force. However if you know of Euler's algorithm, this can make your life much easier.
It sounds like you are working in modular arithmetic. So for $5$ you are supposed to find $x=\frac 11 \pmod 5, y=\frac 12 \pmod 5$, etc. $x$ is pretty easy. For $y$, you need to find $z$ such that $2z=1 \pmod 5$, and so on.