Let's go up a floor

First, let's start with simplifying the left-hand side of the equation. After all, it would be best to get rid of the floor function as early as possible. Using the Basel problem , we know that: $\sum _{ x=1 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 } } } =\frac { { \pi }^{ 2 } }{ 6 }$ For simplicity's sake, I won't prove the entire thing here. Either way, many different proofs for it can be found throughout the internet.

The rest is mere algebra. Knowing that the formula for calculating the infinite sum of a geometric series is ${ S }_{ \infty }=\frac { { a }_{ 1 } }{ 1-r }$ , we can find the value of the second sum on the left: ${ S }_{ \infty }=\frac { { a }_{ 1 } }{ 1-r } =\frac { \frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } } =\frac { \frac { 1 }{ 3 } }{ \frac { 2 }{ 3 } } =\frac { 1 }{ 3 } \cdot \frac { 3 }{ 2 } =\frac { 1 }{ 2 }$
And from there: $\left\lfloor \frac { { \pi }^{ 2 } }{ 6 } +\frac { 1 }{ 2 } \right\rfloor =\left\lfloor \approx 2.14493\dots \right\rfloor =2$ With the left side solved, we can now move to the right side. Let's start by factoring out $n$ : $2=n\left( \sum _{ x=2 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 } } } +\sum _{ x=2 }^{ \infty }{ \frac { 1 }{ { 4 }^{ x } } } \right)$ We could actually try to isolate $n$ at this point by dividing all the known values to one side, but at first, let's solve for what's in the parentheses: $\sum _{ x=2 }^{ \infty }{ \frac { 1 }{ { 4 }^{ x } } } ={ S }_{ \infty }=\frac { \frac { 1 }{ { 4 }^{ 2 } } }{ 1-\frac { 1 }{ 4 } } =\frac { \frac { 1 }{ 16 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 16 } \cdot \frac { 4 }{ 3 } =\frac { 1 }{ 12 } \\ \sum _{ x=2 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 } } } +\sum _{ x=2 }^{ \infty }{ \frac { 1 }{ { 4 }^{ x } } } =\sum _{ x=1 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 } } } -\frac { 1 }{ { 1 }^{ 2 } } +\sum _{ x=2 }^{ \infty }{ \frac { 1 }{ { 4 }^{ x } } } =\frac { { \pi }^{ 2 } }{ 6 } -1+\frac { 1 }{ 12 } =\frac { { \pi }^{ 2 } }{ 6 } -\frac { 11 }{ 12 }$ Now, let's isolate and solve for $n$ : $n=\frac { 2 }{ \frac { { \pi }^{ 2 } }{ 6 } -\frac { 11 }{ 12 } } =\frac { 2 }{ \frac { { \pi }^{ 2 } }{ 6 } \cdot \frac { 2 }{ 2 } -\frac { 11 }{ 12 } } =\frac { 2 }{ \frac { { 2\pi }^{ 2 } }{ 12 } -\frac { 11 }{ 12 } } =\frac { 2 }{ \frac { { 2\pi }^{ 2 }-11 }{ 12 } } =\frac { 2\cdot 12 }{ { 2\pi }^{ 2 }-11 } =\frac { 24 }{ { 2\pi }^{ 2 }-11 } \\ n=\frac { C }{ A{ \pi }^{ A }-B } =\frac { 24 }{ { 2\pi }^{ 2 }-11 } \\ A=2,\quad B=11,\quad C=24\\ A+B+C=2+11+24=\boxed { 37 }$