Divisor Phi Summation

$2016 = 2^5 \cdot 3^2 \cdot 7$ . Since $\varphi$ is a multiplicative function, the answer is simply

$\left( \sum _{ i=0 }^{ 5 }{ \varphi \left( { 2 }^{ i } \right) } \right) \left( \sum _{ i=0 }^{ 2 }{ \varphi \left( { 3 }^{ i } \right) } \right) \left( \sum _{ i=0 }^{ 1 }{ \varphi \left( { 7 }^{ i } \right) } \right) = 2016$

By noting that $\varphi\left(p^k\right) = p^k - p^{k-1}$ ,

$\sum _{ i=0 }^{ n }{ \varphi \left( { p }^{ i } \right) ={ p }^{ n } }$

It can be generalized that

$\sum _{d | k}{\varphi \left( d \right) } = k$