How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$. How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$ . Justify your answer.
Hello, could someone explain to me how to find the invertible elements of a set as large as $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$
I thought about putting together the set and trying each one but it is very big and it can take forever

I will explain how to find the invertible elements of $\mathbb{Z}_n$.
An element $a$ is invertible in $\mathbb{Z}_n$ if and only if therte exists $b\in\mathbb{Z}_n$ such that $a\cdot b=1$, in other words $a\cdot b\equiv 1\pmod{n}$.
$0$ is clearly not invertible.
Observe that if $a$ is not relatively prime with $n$, then this is impossible, as $a\cdot b$ will always be divisible by $\gcd(a,n)$ so this would mean $\gcd(a,n)|1$.
So all elements not relatively prime woth $n$ are not invertible. What about the elements of $\mathbb{Z}_n$ which are relatively prime to $n$? Well let $a$ be one of them. Take the set $\{a,2a,3a,...,(n-1)a\}$. Suppose $ia\equiv ja$ for some $i$ and $j$. This means $a\cdot|i-j|$ is divisible by $n$ so because $a$ is relatively prime to $n$, then $|i-j|$ is divisible by $n$, but $n>|i=j|$ so $|i-j|=0$ so $i=j$.
Thus, all elements of $\{a,2a,3a,...,(n-1)a\}$ are distinct$\pmod{n}$ and there are no elements divisible by $n$ in the above set (because $a\neq 0$ and $a$ and $n$ are coprime), so in the above set there must be an element $\equiv 1\pmod{n}$, thus $a$ is invertible.
So the invertible elements in $\mathbb{Z}_n$ are all the numbers smaller than $n$, which are coprime to $n$. The number of such elements is $\varphi(n)$
