29° 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 $x_1^{{\alpha }_1}+x_2^{{\alpha }_2}+\cdots +x_n^{{\alpha }_n}$ 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 $$\begin{gathered} x_1^{{\alpha }_1}+x_2^{{\alpha }_2}+\cdots +x_n^{{\alpha }_n}=(3^0+3^1+\cdots +3^a)+(4^0+4^1+\cdots +4^b)+(7^0+7^1+\cdots +7^c)= \\ =\frac{3^{a+1}-1}{2}+\frac{4^{b+1}-1}{3}+\frac{7^{c+1}-1}{6}\le \frac{x_{n+1}^{{\alpha }_{n+1}}-1}{2}+\frac{x_{n+1}^{{\alpha }_{n+1}}-1}{3}+\frac{x_{n+1}^{{\alpha }_{n+1}}-1}{6}=x_{n+1}^{{\alpha }_{n+1}}-1. \end{gathered}$$ 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 $x_{n+1}^{{\alpha }_{n+1}}\le m\le x_1^{{\alpha }_1}+x_2^{{\alpha }_2}+\cdots +x_n^{{\alpha }_n}+x_{n+1}^{{\alpha }_{n+1}}$ can be expressed as $m=(m-x_{n+1}^{{\alpha }_{n+1}})+x_{n+1}^{{\alpha }_{n+1}}$,

where $0\le m-x_{n+1}^{{\alpha }_{n+1}}\le x_1^{{\alpha }_1}+x_2^{{\alpha }_2}+\cdots +x_n^{{\alpha }_n}$ has a representation using only $x_1^{{\alpha }_1},\dots ,x_n^{{\alpha }_n}$. We conclude that all integers from $1$ to $x_1^{{\alpha }_1}+\cdots +x_n^{{\alpha }_n}+x_{n+1}^{{\alpha }_{n+1}}$ 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.