Find the Number of elements which are not invertible in the set of integers $\{0, 1, 2, 3, ..... 19\}$ modulo $20$.
Approach:
I have tried finding the elements which satisfies
$k$*(element) mod $20 = 1$ where $k>1;$ the elements which satisfy the equation indeed are the co-primes of $20.$
There are $12$ such elements. Is this correct?
The prime factorization of $20$ is $2^2\times5$, so multiples of $2$ and $5$ are not invertible modulo $20$.
That leaves the following $8$ invertible residues modulo $20$: $1, 3, 7, 9, 11, 13, 17$, and $19.$
Accordingly, Euler's totient function of $20$, $\phi(20)=\phi(4)\times\phi(5)=2\times4=8.$