The 89th question: Given that $\mathrm{n}$ is a given positive integer, and $n \geq 2$, the elements of set $\mathrm{A}$ are all positive integers, the smallest element of $\mathrm{A}$ is 1, and the largest element is $a$, and $7 \times 3^{n}<a<3^{n+2}$, satisfying that for any element $x(x \neq 1)$ in $A$, there exist $y, z, w \in A$ (which can be the same), such that $x=y+z+w$. Find the minimum number of elements in set $A$.
The 89th question: Given that $\mathrm{n}$ is a given positive integer, and $\mathrm{n} \geq 2$, the elements of set $\mathrm{A}$ are all positive integers, the smallest element of $\mathrm{A}$ is 1, and the largest element is $\mathrm{a}$. Moreover, 
$$
7 \times 3^{\mathrm{n}}n+1 \Rightarrow m \geq n+3
$$

Assume $\mathrm{m}=\mathrm{n}+3$. We will use mathematical induction to prove: $\mathrm{a}_{\mathrm{i}}=3^{\mathrm{i}-1}(1 \leq \mathrm{i} \leq \mathrm{m})$.
According to the conditions, it is clear that $a_{2}=a_{1}+a_{1}+a_{1}=3$. Therefore, the equality holds when $n=1, 2$.
Assume that when $\mathrm{i} \leq \mathrm{t}(\mathrm{t}7 \times 3^{\mathrm{n}}$ leads to a contradiction. Therefore, it must be that $\mathrm{a}_{\mathrm{t}+1}=3^{\mathrm{t}}$.
By mathematical induction, we know that $\mathrm{a}_{\mathrm{i}}=3^{\mathrm{i}-1}(1 \leq \mathrm{i} \leq \mathrm{m})$. In particular, $\mathrm{a}=\mathrm{a}_{\mathrm{m}}=3^{\mathrm{m}-1}=3^{\mathrm{n}+2}$, which contradicts $\mathrm{a}<3^{\mathrm{n}+2}$. Thus, $|\mathrm{A}|=\mathrm{n}+3$ cannot satisfy the conditions.
In summary, the smallest value of $|\mathrm{A}|$ that satisfies the conditions is $\mathrm{n}+4$.