If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$ 
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?

My Attempt:
By factoring out powers of $2$, we can write
$$
\begin{align}
K &= 2^{50}\times \left(\frac{1}
{1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\
&= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\
&= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99}
\end{align}
$$
How can I solve for $K$ from here?

Note that this answer is not completely rigorous, but it was too fun to pass up.

$$K^2 = 101 \cdot \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots \cdot 100 \cdot 100}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots \cdot 99 \cdot 101}$$
Now, $$\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots} = \pi / 2$$
(This is known as the Wallis product)
So $K$ is approximately $\sqrt{101 \pi / 2}$ and $\lfloor K \rfloor = 12$
