6. Given a positive integer $k$. If a sequence of subsets of the set of integers $\mathbf{Z}$, $I_{1} \supseteq I_{2} \supseteq \cdots \supseteq I_{k}$, satisfies that for $i=1,2, \cdots, k, 168$ $\in I_{i}$, and for any $x, y \in I_{i}$, we have $x-y \in I_{i}$ $(x, y$ can be equal $)$, then $\left(I_{1}, I_{2}, \cdots, I_{k}\right)$ is called a $k$-chain on $\mathbf{Z}$. How many $k$-chains are there on $\mathbf{Z}$? Please provide a reason.
6. Let $I$ be a non-empty subset of $\mathbf{Z}$ and for any $x, y \in I$, we have $x - y \in I$. Define $I$ as a "good set".

If $I = \{0\}$ or $I = \{a \mid a \in \mathbf{Z}$ and $b \mid a\}$, where $b$ is a positive integer, then $I$ is a good set.

Furthermore, if $I \neq \{0\}$ is a good set, let $b$ be the smallest positive integer in $I$. Then for any $a \in I$, we have $b \mid a$. Otherwise, if $b \nmid a$,

using the division algorithm, there exist unique integers $q, r$ such that
$$
a = bq + r \quad (1 \leq r \leq b-1).
$$

Thus, $r = a - bq \in I$, which contradicts the minimality of $b$. Therefore, $I \neq \{0\}$ is a good set if and only if there exists a positive integer $b$ such that
$$
I = \{a \mid a \in \mathbf{Z} \text{ and } b \mid a\}.
$$

In this case, we denote $I = b\mathbf{Z}$.
Suppose $\left(I_{1}, I_{2}, \cdots, I_{k}\right)$ is a $k$-chain on $\mathbf{Z}$.
Since $168 \in I_{i}$, then $I_{i} \neq \{0\}$.
Suppose $I_{i} = b_{i} \mathbf{Z}$ (where $b_{i}$ is a positive integer and $b_{i} \mid 168$).
Given $168 = 2^3 \times 3 \times 7$, suppose $b_{i} = 2^{x_{i}} 3^{y_{i}} 7^{z_{i}}$.
Since $I_{1} \supseteq I_{2} \supseteq \cdots \supseteq I_{k}$, we have
$b_{i} \mid b_{i+1} (i=1,2, \cdots, k-1)$.
Thus,
$0 \leq x_{1} \leq x_{2} \leq \cdots \leq x_{k} \leq 3$,
$0 \leq y_{1} \leq y_{2} \leq \cdots \leq y_{k} \leq 1$,
$0 \leq z_{1} \leq z_{2} \leq \cdots \leq z_{k} \leq 1$.
It is easy to see that if $1 \leq w_{1} \leq w_{2} \leq \cdots \leq w_{k} \leq a$, let
$t_{1} = w_{1}, t_{2} = w_{2} - w_{1}, \cdots$,
$t_{k} = w_{k} - w_{k-1}, t_{k+1} = a - w_{k}$.
Then $t_{1} + t_{2} + \cdots + t_{k+1} = a$,

where $t_{i} \in \mathbf{N} (i=1,2, \cdots, k+1)$.
Thus, the number of non-negative integer solutions to equation (1) is
$\mathrm{C}_{k+a}^{k} = \mathrm{C}_{k+a}^{a}$.
Therefore, the number of $k$-chains on $\mathbf{Z}$ is
$$
\mathrm{C}_{k+3}^{3} \mathrm{C}_{k+1}^{1} \mathrm{C}_{k+1}^{1} = \frac{(k+1)^{3}(k+2)(k+3)}{6}.
$$