If $\cot \left({\displaystyle \sum _{n=1}^{23}{\cot }^{-1}}\left(1+{\displaystyle \sum _{K=1}^{n}2K}\right)\right)=\frac{p}{q}$  then $\left[\frac{p}{8}\right]$  is, (where [.] denotes the greatest integer function)

$\theta ={\displaystyle \sum _{n=1}^{23}{\cot }^{-1}}\left(1+{\displaystyle \sum _{K=1}^{n}2K}\right)={\displaystyle \sum _{n=1}^{23}{\tan }^{-1}\left(\frac{n+1-n}{1+n\left(n+1\right)}\right)}$

$={\displaystyle \sum _{n=1}^{23}\left({\tan }^{-1}\left(n+1\right)-{\tan }^{-1}n\right)}={\tan }^{-1}24-\frac{\pi }{4}$

$\tan \theta =\frac{23}{25}\Rightarrow \cot \theta =\frac{25}{23}=\frac{p}{q}$

$\left[\frac{p}{8}\right]=\left[3.125\right]=3$