problem: Evaluate $$\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor + .... + \lfloor\sqrt{19}\rfloor$$
solution: We note that if $a^2 \leq n < (a+1)^2$ for some integer $a$, then $a \leq \sqrt{x} < a+1$, so $a$ is the greatest integer less than or equal to $x$. Consequently, we break up our sum into the blocks of integers between consecutive perfect squares:
  
  For $1\leq n \leq 3$, $\lfloor\sqrt{n}\rfloor=1$. There are $3$ values of $n$ in this range.
  For $4\leq n\leq 8$, $\lfloor\sqrt{n}\rfloor=2$. There are $5$ values of $n$ in this range.
  For $9\leq n \leq 15$, $\lfloor\sqrt{n}\rfloor=3$. There are $7$ values of $n$ in this range.
  For $16\leq n \leq 19$, $\lfloor\sqrt{n}\rfloor=4$. There are $4$ values of $n$ in this range.
  
  Consequently, our total sum is $3\cdot1+5\cdot2+7\cdot3+4\cdot 4= \boxed{50}$.