Annual Harvard-MIT Mathematics Tournament

Solution:

For $n=12=2^{2} \cdot 3^{1}$,
$$\phi(12)=2(2-1)(3-1)=4, \quad \sigma(12)=(1+2+4)(1+3)=28, \quad \tau(12)=(2+1)(1+1)=6$$

For $n=2007=3^{2} \cdot 223$,
$$\phi(2007)=3(3-1)(223-1)=1332, \quad \sigma(2007)=(1+3+9)(1+223)=2912, \quad \tau(2007)=(2+1)(1+1)=6$$

For $n=2^{2007}$,
$$\phi\left(2^{2007}\right)=2^{2006}, \quad \sigma\left(2^{2007}\right)=\left(1+2+\cdots+2^{2007}\right)=2^{2008}-1, \quad \tau\left(2^{2007}\right)=2007+1=2008$$