Simple Simplification III

$\quad \quad \left\lfloor \frac { { 2 }^{ 31 }{ +3 }^{ 31 } }{ { 2 }^{ 29 }+{ 3 }^{ 29 } } \right\rfloor \\ \\ =\quad \left\lfloor \frac { { 3 }^{ 31 }(1+{ \frac { 2 }{ 3 } }^{ 31 }) }{ { 3 }^{ 29 }(1+{ \frac { 2 }{ 3 } }^{ 29 }) } \right\rfloor \\ \\ =\quad \left\lfloor \frac { 9(1+{ \frac { 2 }{ 3 } }^{ 31 }) }{ (1+{ \frac { 2 }{ 3 } }^{ 29 }) } \right\rfloor$

Now in the floor function, we can see that $2/3$ is less than 1. So, $2/3$ raised to the power $29$ and $31$ is also less than 1. But,

$\quad \quad \quad { (\frac { 2 }{ 3 } })^{ 29 }\quad >{ \quad (\frac { 2 }{ 3 } })^{ 31 }\\ \Rightarrow \quad 1+{ (\frac { 2 }{ 3 } })^{ 29 }\quad >\quad 1+({ \frac { 2 }{ 3 } })^{ 31 }\\ \Rightarrow \quad \frac { 1+{ (\frac { 2 }{ 3 } })^{ 31 } }{ 1+{ (\frac { 2 }{ 3 } })^{ 29 } } \quad <\quad 1\\ \\ \Rightarrow \quad 9(\frac { 1+{ (\frac { 2 }{ 3 } })^{ 31 } }{ 1+{ (\frac { 2 }{ 3 } })^{ 29 } } )\quad <\quad 9\\ \\ \Rightarrow \quad \left\lfloor 9(\frac { 1+{ (\frac { 2 }{ 3 } })^{ 31 } }{ 1+{ (\frac { 2 }{ 3 } })^{ 29 } } ) \right\rfloor \quad =\quad 8$

Which is the required value!!!

$\dfrac{2^{31}+2^{31}}{2^{29}+3^{29}}$

$= \dfrac{(2^{29}+3^{29})(2^2+3^2)-4(2^{29}+3^{29})-5 \cdot 2^{29}}{2^{29}+3^{29}}$

$= 13-4- \dfrac{5 \cdot 2^{29}}{3^{29}+2^{29}}$

It can be easily seen that $\dfrac{5 \cdot 2^{29}}{3^{29}+2^{29}}$ is less that $1$

it follows that $8 < \dfrac{2^{31}+2^{31}}{2^{29}+3^{29}} < 9$

So the answer is $\boxed{8}$

$\begin{aligned} \frac{2^{31}+3^{31}}{2^{29}+3^{29}} & = \frac{4(2^{29})+9(3^{29})}{2^{29}+3^{29}} \\ & = 4 + \frac{5(3^{29})}{2^{29}+3^{29}} \\ & = 4 + \frac{\color{#3D99F6}{3^{29}} + 4(3^{29})}{2^{29}+3^{29}} \quad \quad \quad \color{#3D99F6}{\text{We note that }3^{29} = (2+1)^{29} = 2^{29} + 29(2^{28})+... > 4(2^{29})} \\ & = 4 + \frac{\color{#3D99F6}{3^{29} - 4(2^{29}) + 4(2^{29})} + 4(3^{29})}{2^{29}+3^{29}} \\ & = 4 + 4 + \color{#D61F06}{\frac{3^{29} - 4(2^{29})}{2^{29}+3^{29}}} \quad \quad \color{#D61F06}{\text{Since }\frac{3^{29} - 4(2^{29})}{2^{29}+3^{29}}<1} \\ \Rightarrow \left \lfloor \frac{2^{31}+3^{31}}{2^{29}+3^{29}} \right \rfloor & = \boxed{8} \end{aligned}$