Olimpiada Matemática del Cono Sur

Consider the powers of $3$, $4$ and $7$ in increasing order $x_1^{\alpha_1} \le x_2^{\alpha_2} \le x_3^{\alpha_3} \le \dots$
We will prove, by induction on $n$, that it is possible to represent all the integers from $1$ to $x_1^{\alpha_1} + x_2^{\alpha_2} + \dots + x_n^{\alpha_n}$ in the desired way using only the powers $x_1^{\alpha_1}, x_2^{\alpha_2}, \dots, x_n^{\alpha_n}$. It is clear that this is true for $n = 1, 2, 3$. Assuming the property holds for $n \ge 3$, we will prove that it is valid for $n + 1$.
If the powers $x_1^{\alpha_1}, x_2^{\alpha_2}, \dots, x_n^{\alpha_n}$ are $\{3^0, 3^1, \dots, 3^a\} \cup \{4^0, 4^1, \dots, 4^b\} \cup \{7^0, 7^1, \dots, 7^c\}$, we have that
$$x_1^{\alpha_1} + x_2^{\alpha_2} + \dots + x_n^{\alpha_n} = (3^0 + 3^1 + \dots + 3^a) + (4^0 + 4^1 + \dots + 4^b) + (7^0 + 7^1 + \dots + 7^c) = \\ = \frac{3^{a+1}-1}{2} + \frac{4^{b+1}-1}{3} + \frac{7^{c+1}-1}{6} \le \frac{x_{n+1}^{\alpha_{n+1}}-1}{2} + \frac{x_{n+1}^{\alpha_{n+1}}-1}{3} + \frac{x_{n+1}^{\alpha_{n+1}}-1}{6} = x_{n+1}^{\alpha_{n+1}} - 1.$$
Then, by the induction assumption, all the positive integers smaller than $x_{n+1}^{\alpha_{n+1}}$ have a representation using only the powers $x_1^{\alpha_1}, x_2^{\alpha_2}, \dots, x_n^{\alpha_n}$. In addition, an integer $m$ such that $x_{n+1}^{\alpha_{n+1}} \le m \le x_1^{\alpha_1} + x_2^{\alpha_2} + \dots + x_n^{\alpha_n} + x_{n+1}^{\alpha_{n+1}}$ can be expressed as $m = (m - x_{n+1}^{\alpha_{n+1}}) + x_{n+1}^{\alpha_{n+1}}$,

where $0 \le m - x_{n+1}^{\alpha_{n+1}} \le x_1^{\alpha_1} + x_2^{\alpha_2} + \dots + x_n^{\alpha_n}$ has a representation using only $x_1^{\alpha_1}, \dots, x_n^{\alpha_n}$. We conclude that all integers from $1$ to $x_1^{\alpha_1} + \dots + x_n^{\alpha_n} + x_{n+1}^{\alpha_{n+1}}$ have a representation of the desired form using only the powers $x_1^{\alpha_1}, \dots, x_n^{\alpha_n}, x_{n+1}^{\alpha_{n+1}}$, which completes the induction.