Let $n = 2^{k_1} × 3^{k_2} × 5^{k_3}$.
Show that $\phi (n) = \frac{4}{15}n$.
We have
$$ \begin{align} \phi(n) & = 2^{k_1 - 1}(2-1) \times 3^{k_2 -1 }(3-1) \times 5^{k_3 - 1}(5-1) \\ \\ & = \frac{1\times 2 \times 4}{2 \times 3 \times 5}n \\ \\ & = \frac{4}{15}n \end{align}$$