I wonder if you can help me with this question I am being dealing with. My line of thinking was this:
I know that the sequence is of length $n$, so I divided it into $n$ cells. $$x_1+x_2+x_3+\dotsb +x_n=tk,\quad1\le t\le n.$$ So its like dividing $tk$ white balls into $n$ cells. But each cell has to hold at least $1$ white ball so the problem changes to divide $tk-n$ white balls into $n$ cells.
This is my answer but I don't know if I have counted options twice. On top of that I got an open answer (with operator sum) but I need a closed answer (without the sum). This is what I achieved so far. Hope anyone can help or give me a different way of thinking about this problem.
$$\sum_{t=1}^{n}{n+(tk-n)-1 \choose tk-n}=\sum_{t=1}^{n}{tk-1 \choose tk-n}$$
ok guys so i got the answer and i would like to know what do you think of it. so let's look at all of the sequences taken from $\{1,2,...,k\}$ that their length is $n-1$. there are $k^{n-1}$ such options. now if we look at any sequence as a sum of it's elements, we will see that the modulo $k$ of the sum can variate from $0$ to $k-1$. so there is only one way to add the last number in the sequence to make it a sequence from length $n$ that is divisible by $k$ . so the final answer is: $k^{n-1}$ ways