Euler's totient function of 18 - phi(18) I am trying to find the phi(18). Using an online calculator, it says it is 6 but im getting four. 

The method I am using is by breaking 18 down into primes and then multiplying the phi(primes)
$$=\varphi  (18)$$
$$=\varphi  (3) \cdot \varphi(3) \cdot \varphi(2)$$
$$= 2 \cdot 2 \cdot 1$$
$$= 4$$
 A: Remember that you need to determine the prime factorization of $18$.  That is, $18 = 3^2 \cdot 2$.  Since $18 = 3^2 \cdot 2$, we have
$$\begin{aligned}
\varphi(18) = \varphi(3^2) \cdot \varphi(2) &= (3^2 - 3)(2 - 1) = 6
\end{aligned}$$
So in general, if $k = p_1^{n_1}p_2^{n_2}\cdots p_s^{n_s}$, then we have
$$\varphi(k) = \varphi(p_1^{n_1})\varphi(p_2^{n_2})\cdots \varphi(p_s^{n_s}) = (p_1^{n_1} - p_1^{n_1 - 1})(p_2^{n_2} - p_2^{n_2 - 1})\cdots (p_s^{n_s} - p_s^{n_s - 1})$$
A: The phi function is multiplicative, but not completely multiplicative: Thus if $a, b$ are relatively prime, we have that
$$\varphi(ab) = \varphi(a) \varphi(b)$$
but this is not necessarily true if $a$ and $b$ have a common prime factor. In particular, it's true that
$$\varphi(18) = \varphi(2) \varphi(9)$$
but not $\varphi(2) \varphi(3)^2$. 
A: Your multiplicative property is not necessarily true when the two numbers you're multiplying share a common factor. Here's the general formula: given $N = p_1^{q_1}p_2^{q_2}\cdots p_n^{q_n}$ (where $p_1, \ldots, p_n$ are distinct primes), we can find that 
$$
\phi(N) = N\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots\left(1-\frac{1}{p_n}\right)
$$
In the case of $18 = 2 \cdot 3^2$, this gives:
$$
\phi(18) = 18\left(\frac{1}{2}\right)\left(\frac{2}{3}\right) = 6
$$
I leave the proof of this as an exercise (hint: consider (and enumerate) those numbers divisible by $p_1, p_2, \ldots, p_n$).  
A: By definition, $\varphi(18)$ is the number of elements in the set 
$$\{n : 1\leq n \leq 17, \text{ with } \gcd(n,18)=1\}=\{1,5,7,11,13,17\}.$$
Thus, $\varphi(18)=6$. Similarly, $\varphi(9)$ is the number of elements in the set
$$\{n : 1\leq n \leq 8, \text{ with } \gcd(n,9)=1\}=\{1,2,4,5,7,8\},$$
so $\varphi(9)=6$, and not $4$.
