I cannot get the correct answer of this problem. The answer is 445 but I got 435.
May be I didn't understand what the last sentence meant.
First, make a simple conversion
$\displaystyle \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{{n^2} + 8n + 10}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{n\left( {n + 9} \right) + 10 - n}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{30} {\left\lfloor {n + \frac{{10 - n}}{{n + 9}}} \right\rfloor } =$
$\displaystyle = \sum\limits_{n = 1}^{30} n + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = 465 + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } .$
$\displaystyle \frac{{10 - n}}{{n + 9}} = 0 \Leftrightarrow n = 10$
Therefore, we consider two cases:
1) if $\displaystyle n = \left\{ {1,{\text{ }}2,{\text{ }} \ldots ,{\text{ }}10} \right\}$ then $\displaystyle \left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor = 0$.
2) if $\displaystyle n = \left\{ {11,{\text{ 1}}2,{\text{ }} \ldots ,{\text{ }}30} \right\}$ then $\displaystyle \left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor = - 1$.
So we have
$\displaystyle \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{10} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } + \sum\limits_{n = 11}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = 0 + \left( { - 20} \right) = - 20$.
Finally we have
$\displaystyle \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{{n^2} + 8n + 10}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{30} n + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = 465 + \left( { - 20} \right) = 445.$