(12) Let $[x]$ denote the greatest integer not exceeding $x$, $a_{k}=\left[\frac{2009}{k}\right]$, $k=1$,
$2, \cdots, 100$, then the number of different integers among these 100 integers is $\qquad$
(12) 69 Hint: When $01$, $[a+b]>[a]$.
$$
\begin{array}{l}
\frac{2009}{k}-\frac{2009}{k+1}=\frac{2009}{k(k+1)}, \\
\frac{2009}{k}=\frac{2009}{k+1}+\frac{2009}{k(k+1)} .
\end{array}
$$

When $k \leqslant 44$, $\frac{2009}{k(k+1)}>1$, and when $k \geqslant 45$, $\frac{2009}{k(k+1)}<1$, so when $k \leqslant 44$, $a_{1}, \cdots, a_{44}$ are all different. $a_{44}=45, a_{45}=44, a_{99}=20$, $a_{100}=20$. When $k \geqslant 45$, $a_{k}=a_{k+1}$ or $a_{k}=a_{k+1}+1$, so $a_{45}, \cdots, a_{100}$ take all the integers from 20 to 44. Therefore, $a_{1}, a_{2}, \cdots, a_{100}$ contain a total of $44+25=$ 69 (numbers) different integers.