Compute $\varphi (180)$ where $\varphi$ is the Euler totient fuction

Question

So I was given this question. Compute $\varphi (180)$ where $\varphi$ is the Euler totient fuction.

Here is my attempted solution:

$\varphi (180) = ?$ $m = 180 = 2^2 \cdot 5 \cdot 9$, $p_1 = 2, p_2 = 5, p_3 = 9, n = 5$

$\varphi (180) = \sum_{k=0}^5 (-1)^k = \frac{(180)}{i\varepsilon S\prod P_i} = \frac{(180)}{1} - \frac{(180)}{p_1} - \frac{(180)}{P_2} - \frac{(180)}{P_3} + \frac{(180)}{P_1P_2} + \frac{(180)}{P_1P_3} + \frac{(180)}{P_2P_3} - \frac{(180)}{P_1P_2P_3} $

This is all i really know, basically the general process on how to start the problem off but I get confused on how to go about it with subsets, etc.

Answer 1

Hint: Use co-prime multiplicativity, i.e. the property that $\varphi(pq) = \varphi(p)\varphi(q)$ when $(p, q) = 1$ to break it into individual prime factors.

It should then be easy to calculate each individual $\varphi(p^k)$.

Answer 2

The problem with your solution is that you are not removing the numbers having $GCD$ other that $2,5,9$ with $180$. Proceeding with your method, you should take into account every divisor of $180$ and then apply the same procedure. This will give you the correct answer.

Also it can be solved by using multiplicativity of totient function.

Answer 3

$$\phi (n)=n×(1-1/p_1)(1-1/p_2)..... (1-1/p_k)$$ Where $n=(p_1)^{n_1}(p_2)^{n_2} (p_3)^{n_3}....(p_k)^{n_k}$.