29. For a positive integer $k$, we write
$$
(1+x)(1+2 x)(1+3 x) \cdots(1+k x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{k} x^{k},
$$
where $a_{0}, a_{1}, \cdots, a_{k}$ are the coefficients of the polynomial. Find the smallest possible value of $k$ if $a_{0}+a_{1}+a_{2}+\cdots+a_{k-1}$ is divisible by 2005 .
29. Ans: 401
By comparing the coefficients of $x^{k}$, it is easy to see that $a_{k}=k!$. Letting $x=1$, we get
$$
a_{0}+a_{1}+\cdots+a_{k}=(1+1)(1+2)(1+3) \cdots(1+k)=(k+1)!
$$

Therefore, $a_{0}+a_{1}+\cdots+a_{k-1}=(k+1)!-k!=k \cdot k!$. Note that $2005=5 \times 401$, where 401 is a prime number. It is easy to see that a prime number $p$ divides $k$ ! only if $k>p$. Therefore, the smallest possible value of $k$ is 401 .