Find the sum of the digits of the smallest value of $k$ if $(a_0 + a_1 +$ ... $+ a_{k-1})$ is divisible by $2005$. 
For a positive integer $k$, we have:- $(1 + x)(1 + 2x)(1 + 3x)$ ... $(1 + kx) = a_0 + a_1x + a_2x^2 + $ ... $ + a_kx^k$. Where $a_0,a_1,$ ... $,a_k$ are the coefficients of the polynomial. Find the sum of the digits of the smallest value of $k$ if $(a_0 + a_1 +$ ... $+ a_{k-1})$ is divisible by $2005$.

What I Tried: I have no idea where to start. I was able to guess, that the expression $(1 + x)(1 + 2x)$ ... $(1 + kx)$ consists of a $(+1)$ term and a $(k!x^k)$ term , and I can guess $a_0 = 1$ , $a_k = k!$ , and this might given an idea that $a_1 = 1!$ , $a_2 = 2!$ and so on, but is this enough.
Suppose, after solving this part, I need to find for which $(k-1)$ I will have that :-
$$\rightarrow 0! + 1! + ... + (k - 1)! = 2005m$$
This is easy, because as far as I can see, $2005 = 5 * 401$ and $401$ is prime , so $k$ has to be $402$ in order to be minimum.
I have a doubt in the $1$st part, can anyone help me?
 A: $(1+x)(1+2x) = 1+3x+2x^2$ disproves your claim that $a_1 = 1!$.
Alternatively, consider $f(1)$. We have:
$$f(1) = (1+1)(1+2)\dots(1+k) = a_0+a_1+\dots+a_k = (k+1)!$$
and we want $(k+1)! - a_k = (k+1)! - k! = k(k!)$ to be divisible by $2005$.
By a similar analysis to yours, the minimum $k$ is $401$.
A: Hint:
$$a_0+a_1+a_2+\dots +a_{k} = f(1) = (k+1)! $$
A: Just note that if you put $P(x)=(1+x)(1+2x)...(1+kx)$ then the value of $a_{0}+a_{1}+...+a_{k-1}$ is actually equal to $P(1)-a_{k}=P(1)-k!=(k+1)!-k!$ Now if $(k+1)!\equiv k! (mod401)$ for a $400\geqslant k$ (Note: It is obviously divisible by $5$ for $k\geqslant 5$ so it suffices that it's divisible by $401$)
Then because $401$ is a prime and $k!,(k+1)!$ are both relatively prime to it, $k+1\equiv 1 (mod401)$ and hence the least such $k$ is $401$ which is a contradiction so none of the numbers less than $401$ satisfy the condition. $401$ itself clearly satisfies the condition so the final answer is $5$.
A: Here there are some hints. Let's call the polynomial $p(x) $.

*

*In terms of $a_i$, how do you express $p(1) $?

*What is the coefficient of $x^k$ in $p(x) $?

After having solved this part, you are almost done... You noticed that 401 is prime :)
Good luck!
