For the 74th question, for real $\alpha>1$, define the infinite sequence $\left\{a_{n}\right\}_{n \in \mathbb{N}}: a_{n}=\left[\alpha^{n}\right]$. If for any $m \in \mathbb{Z}^{+}$, there exist a positive integer $k$ and $k$ natural numbers $0 \leq i_{1}<i_{2}<\ldots<i_{k}$, such that $m=\sum_{j=1}^{k} a_{i j}$, find all possible values of $\alpha$.

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The 74th question, for real $\alpha>1$, defines the infinite sequence $\left\{a_{n}\right\}_{n \in \mathbb{N}}: a_{n}=\left[\alpha^{n}\right]$. If for any $m \in \mathbb{Z}^{+}$, there exist a positive integer $k$ and $k$ natural numbers $0 \leq i_{1}<i_{2}<\ldots<i_{k}$, such that $m=\sum_{j=1}^{k} a_{i j}$, try to find all possible values of $\alpha$.
The 74th question:
Solution: If $\alpha$ satisfies the condition, then for any positive integer $\mathrm{n},\left[\alpha^{\mathrm{n}}\right]-1$ can only be expressed as the sum of some terms in $\left[\alpha^{\mathrm{i}}\right](0 \leq \mathrm{i} \leq \mathrm{n}-1)$, hence
$$
\alpha^{\mathrm{n}}-22(\sqrt[4]{13})^{4}-(\sqrt[4]{13})^{5}>1$, thus $2\left[\alpha^{n}\right]-\left[\alpha^{n+1}\right] \geq\left[2 \alpha^{n}-\right.$ $\left.\alpha^{\mathrm{n}+1}\right]-1 \geq 0$. According to the induction hypothesis:
$\left[\alpha^{\mathrm{n}+1}\right] \leq\left[\alpha^{\mathrm{n}}\right]+\left[\alpha^{\mathrm{n}}\right] \leq\left[\alpha^{\mathrm{n}}\right]+1+\sum_{\mathrm{i}=0}^{\mathrm{n}-1}\left[\alpha^{\mathrm{i}}\right]=1+\sum_{\mathrm{i}=0}^{\mathrm{n}}\left[\alpha^{\mathrm{i}}\right]$
In summary, the lemma is proved!
Now, let's return to the original problem.
We use mathematical induction on the natural number $\mathrm{n}$ to prove: when the integer $\mathrm{m}$ satisfies $1+\sum_{\mathrm{i}=0}^{\mathrm{n}-1}\left[\alpha^{\mathrm{i}}\right] \leq \mathrm{m}<1+\sum_{\mathrm{i}=0}^{\mathrm{n}}\left[\alpha^{\mathrm{i}}\right]$, then $\mathrm{m}$ can be expressed as the sum of some terms in $\left[\alpha^{\mathrm{i}}\right](0 \leq \mathrm{i} \leq \mathrm{n})$.
Notice that $\left[\alpha^{0}\right]=1,\left[\alpha^{1}\right]=1$. Hence, when $\mathrm{n}=0, 1$, the conclusion holds.
Assume the conclusion holds for $\mathrm{n}$.
We consider the case of $\mathrm{n}+1$. According to the lemma, at this time $\left[\alpha^{\mathrm{n}+1}\right] \leq 1+\sum_{\mathrm{i}=0}^{\mathrm{n}}\left[\alpha^{\mathrm{i}}\right] \leq \mathrm{m}<1+\sum_{\mathrm{i}=0}^{\mathrm{n}+1}\left[\alpha^{\mathrm{i}}\right]$. According to the lemma:
$$
\mathrm{m}-\left[\alpha^{\mathrm{n}+1}\right] \leq \sum_{i=0}^{\mathrm{n}+1}\left[\alpha^{\mathrm{i}}\right]-\left[\alpha^{\mathrm{n}+1}\right] \leq \sum_{i=0}^{\mathrm{n}}\left[\alpha^{i}\right]
$$

Thus, according to the induction hypothesis, $\mathrm{m}-\left[\alpha^{\mathrm{n}}\right]$ can be expressed as the sum of some terms in $\left[\alpha^{\mathrm{i}}\right](0 \leq \mathrm{i} \leq \mathrm{n})$, hence $\mathrm{m}$ can be expressed as the sum of some terms in $\left[\alpha^{\mathrm{i}}\right](0 \leq \mathrm{i} \leq \mathrm{n}+1)$.
In summary, the required $\alpha \in(1, \sqrt[4]{13}) \cup\{2\}$.
Note 1: When $\alpha \in(1,2)$, $a_{0}=a_{1}=1$. When $n \geq 2$, we have $a_{n} \leq \sum_{i=1}^{n-1} a_{i}+1$, hence it is easy to know $a_{n} \leq$ $3 \cdot 2^{\mathrm{n}-2}$.
$$
\text { Also } \alpha^{n}<\left[\alpha^{n}\right]+1=a_{n}+1 \Rightarrow \alpha<\left(3 \cdot 2^{n-2}+1\right)^{\frac{1}{n}}, \text { thus } \alpha<\min _{\substack{n \geq 3 \\ n \in \mathbb{Z}}}\left(3 \cdot 2^{n-2}+1\right)^{\frac{1}{n}}=\left(3 \cdot 2^{4-2}+\right.
$$
$1)^{\frac{1}{4}}=\sqrt[4]{13}$
Note 2: For the non-decreasing sequence of positive integers $\mathrm{a}_{1} \leq \mathrm{a}_{2} \leq \mathrm{a}_{3} \leq \ldots$, then
(1) When and only when $a_{1}=1$, and $a_{n} \leq 1+\sum_{i=1}^{n-1} a_{i}(n \geq 2)$, every positive integer can be expressed as the sum of some terms in this sequence;
(2) When and only when $a_{1}=1$, and $a_{n}=1+\sum_{i=1}^{n-1} a_{i}(n \geq 2)$, every positive integer can be uniquely expressed as the sum of some terms in this sequence.
This proposition is the theoretical basis of positional notation.