Show that $\phi (10^m) = 4 (10^{m−1})$ where $m$ is a natural number.
Since $\gcd(2^m, 5^m)=1$, we can use the multiplicity of $\phi$,
$$ \begin{align} \phi(10^m) & = \phi(2^m5^m)\\ \\ & = \phi(2^m) \times \phi(5^m) \\ \\ & = 2^{m-1}(2-1) \times 5^{m-1}(5-1) \\ \\ & = 4(10^{m-1}) \end{align} $$