How many elements are in the invertible set $\mathbb{Z}_n$?

Question

My question is directly, how many elements are in the invertible set $Z_{35}$?

It's my understanding that for any $Z_n$, if $n$ is prime, then the number of invertible elements is equal to $n-1$. In addition, all elements that are invertible satisfy the formula $\gcd(x,n) = 1$. Thus, for $n = 35$, which is composed of two primes, the number of invertible elements should be $n-3$, or $32$. The non-invertible elements namely $0, 5, 7$.

Will someone please show me where I my logic is incorrect. This was a Coursera question that I missed four times, but I can't seem to figure out why unless I am misunderstanding something or missing something very obvious.

Answer 1

What is $\gcd(10,35)$? What is $\gcd(14,35)$? Can you begin to see from these examples which other non-invertible elements of $\mathbb{Z}_{35}$ you have been missing?

Answer 2

it is called as Euler-phi function.$\Phi (n)=|\{1\leq a <n|(a,n)=1\}|$.

And $\Phi(p)=p-1$ for prime numbers and it is multiplicative i.e if $(m,n)=1$ then $\Phi(mn)=\Phi(m)\Phi(n)$.When you know this,it is easy to compute. $$\Phi(35)=\Phi(7)\Phi(5)=6\cdot4=24$$

Answer 3

The formula for Euler's $\phi$ function answers that.