Question 232, Set $P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a 、 b 、 c\right.$ are positive integers not exceeding 6 $\}$, if $x_{1} 、 x_{2} 、 \ldots 、 x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$.
Question 232, Solution: On the one hand, $7^{3}+7^{2}+7+1$, $7^{3}+7^{2}+7+2$, $7^{3}+7^{2}+7+3$, $7^{3}+7^{2}+7+4$, $7^{3}+7^{2}+7+5$, $7^{3}+7^{2}+7+6$, are six elements in set $P$ that form an arithmetic sequence, so $\mathrm{n}_{\max } \geq 6$.

On the other hand, all elements in set $P$ can be represented as four-digit numbers in base 7, with all digits being non-zero and the leading digit being 1. We can prove that any seven different elements in set $P$ do not form an arithmetic sequence.
Otherwise, if there exist seven numbers $a_{1}$, $a_{2}$, ..., $a_{7} \in P$ that form an arithmetic sequence with a common difference $d$. Note that $1 + 6d \leq a_{1} + 6d = a_{7} < 7^{4}$, so $d < 7^{4}$. Let $7^{\alpha} \| d$, then $0 \leq \alpha \leq 3$. Thus, the seven numbers $a_{1}$, $a_{1}+d$, $a_{1}+2d$, ..., $a_{1}+6d$ when represented in base 7, their $(\alpha+1)$-th digit from the right forms a complete residue system modulo 7, and thus there must be a number whose $(\alpha+1)$-th digit is 0, which means this number does not belong to $P$, leading to a contradiction.
In summary, the maximum value of $\mathrm{n}$ is 6.
Note: Similarly, we can prove the following proposition: Given a prime number $\mathrm{p}$, there exists a set $S$ such that any $\mathrm{p}$ different elements in set $S$ do not form an arithmetic sequence, and for any $a \in S$, there exist $a_{1}=a$, $a_{2}$, $a_{3}$, ..., $a_{p-1}$ as different elements in set $S$ that form an arithmetic sequence.

Proof: Define the set $S=\{x \mid x \in N, x$'s $p$-ary representation does not contain the digit $p-1\}$. We will prove that this set $S$ satisfies the given conditions.

First, for any $a \in S$, let the $p$-ary representation of $a$ be $a=\left(\overline{n_{k} n_{k-1} \ldots n_{0}}\right)_{p}$. Take $a_{i}=(i-1)p^{k+1}$, where $i=1$, $2$, ..., $p-1$, then $\left\{a_{i}\right\}_{1 \leq i \leq p-1}$ forms an arithmetic sequence, and in the $p$-ary representation of $a_{i}$, the leading digit is $i-1$, and the other digits are $n_{k}$, $n_{k-1}$, ..., $n_{0}$, none of which are $p-1$, so $a_{i} \in P$.
Next, we prove that set $S$ does not contain $p$ numbers that form an arithmetic sequence.
Otherwise, if there exist $p$ numbers $a_{1}$, $a_{2}$, ..., $a_{p} \in S$ that form an arithmetic sequence with a common difference $d$. Let $p^{\alpha} \| d$, $\alpha \in N$. Then the $p$ numbers $a_{1}$, $a_{1}+d$, $a_{1}+2d$, ..., $a_{1}+(p-1)d$ when represented in base $p$, their $(\alpha+1)$-th digit from the right forms a complete residue system modulo $p$, and thus there must be a number whose $(\alpha+1)$-th digit is $p-1$, which means this number does not belong to $S$, leading to a contradiction.
In summary, the set $S$ defined above satisfies the given conditions.