Multiply by thee is hard

If $n=3^mq$ where $3\not | q$ , then $\sigma_0(n)=\sigma_0(3^m)\sigma_0(q)=(m+1)\sigma_0(q)=8$ , so that $m+1$ can be 1,2,4 or 8. Now $\sigma_0(3n)=\sigma_0(3^{m+1})\sigma_0(q)=\frac{8(m+2)}{m+1}$ , which can be 16, 12, 10, or 9. The answer is $\boxed{47}$

If $\displaystyle { \sigma }_{ 0 }\left( n \right) =8$ ,

then it has to be expressed as $\displaystyle { p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }$ , $\displaystyle { p }_{ 1 }^{ 3 }{ p }_{ 2 }$ or $\displaystyle { p }_{ 1 }^{ 7 }$ .

Scenario 1: $\displaystyle n = { p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }$

Case 1: ${ p }_{ 1 }=3$

Since $\displaystyle n = { p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }$

So $\displaystyle 3n=3{ p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }\\ ={ p }_{ 1 }^{ 2 }{ p }_{ 2 }{ p }_{ 3 }\Longrightarrow 3\times 2\times 2=12\quad factors$

Case 2: ${ p }_{ 2 }=3$

Since $\displaystyle n = { p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }$

So $\displaystyle 3n=3{ p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }\\ ={ p }_{ 2 }^{ 2 }{ p }_{ 1 }{ p }_{ 3 }\Longrightarrow 3\times 2\times 2=12\quad factors$

Case 3: ${ p }_{ 3 }=3$

Since $\displaystyle n = { p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }$

So $\displaystyle 3n=3{ p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }\\ ={ p }_{ 3 }^{ 2 }{ p }_{ 1 }{ p }_{ 2 }\Longrightarrow 3\times 2\times 2=12\quad factors$

Case 4: ${ p }_{ 1,2,3 }\neq 3$

Since $\displaystyle n = { p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 }$

So $\displaystyle 3n=3{ p }_{ 1 }{ p }_{ 2 }{ p }_{ 3 } \Longrightarrow 2\times 2\times 2\times 2=16\quad factors$

Scenario 2: $\displaystyle n= { p }_{ 1 }^{ 3 }{ p }_{ 2 }$

Case 1: ${ p }_{ 1 }=3$

Since $\displaystyle n= { p }_{ 1 }^{ 3 }{ p }_{ 2 }$

So $\displaystyle 3n={ 3p }_{ 1 }^{ 3 }{ p }_{ 2 }\\ ={ p }_{ 1 }^{ 4 }{ p }_{ 2 } \Longrightarrow 5\times 2=10\quad factors$

Case 2: ${ p }_{ 2 }=3$

Since $\displaystyle n= { p }_{ 1 }^{ 3 }{ p }_{ 2 }$

So $\displaystyle 3n={ 3p }_{ 1 }^{ 3 }{ p }_{ 2 }\\ ={ p }_{ 1 }^{ 3 }{ p }_{ 1 }^{ 2 } \Longrightarrow 4\times 3=12\quad factors$

Case 3: ${ p }_{ 1,2 } \neq 3$

Since $\displaystyle n= { p }_{ 1 }^{ 3 }{ p }_{ 2 }$

$\displaystyle 3n={ 3p }_{ 1 }^{ 3 }{ p }_{ 2 }\Longrightarrow 2\times 4\times 2=16\quad factors$

Scenario 3: $\displaystyle n= { p }_{ 1 }^{ 7 }$

Case 1: ${ p }_{ 1 }=3$

Since $\displaystyle n= { p }_{ 1 }^{ 7 }$

$\displaystyle 3n=3{ p }_{ 1 }^{ 7 }\\ ={ p }_{ 1 }^{ 8 }\Longrightarrow 9\quad factors$

Case 2: ${ p }_{ 1 } \neq 3$

Since $\displaystyle n= { p }_{ 1 }^{ 7 }$

$\displaystyle 3n=3{ p }_{ 1 }^{ 7 }\\ ={ 3p }_{ 1 }^{ 7 }\Longrightarrow 8\times 2=16\quad factors$

That's all the cases so $12+16+10+9=\boxed { 47 }$