XXVII Olimpiada Matemática Rioplatense

For every positive integer $m$, we know that $m$ and $S(m)$ have the same remainder $r$ modulo $3$. Then, $m+2S(m)$ has the same remainder as $3r$ and so, it is a multiple of $3$. Therefore, if a number $n$ is rioplatense, then it is a multiple of $3$.

Now we are going to show that every multiple of $3$ is rioplatense. We will prove by induction on $k$ that every integer $n$ divisible by $3$ with at most $k$ digits can be written as $n=m+2S(m)$ for an integer $m$ with at most $k$ digits.

For $k=1$, the result is immediate, since $3=1+2S(1)$, $6=2+2S(2)$ and $9=3+2S(3)$. Assume the result holds for $k\ge 1$ and consider an integer $n$ multiple of $3$ with $k+1$ digits. We have $10^k+2\le n<10^{k+1}$, since $10^k+2=10\cdots 02$ is the smallest multiple of $3$ with $k+1$ digits. Let us call $N=10^k+2$. When dividing $n$ by $N$, we obtain a quotient $q$ and a remainder $r$. Note that $1\le q\le 9$, since $n<10^{k+1}<10N$. On the other hand, $r<10^k+2$ and it is a multiple of $3$, since $r=n-Nq$ and both $n$ and $N$ are multiples of $3$, which implies that $r\le 10^k-1$ and so, it has at most $k$ digits. By the induction assumption, there exists $t=\overline{a_{k-1}\cdots a_1{a}_0}$ such that $r=t+2S(t)$.

Take $m=\overline{a_k{a}_{k-1}\cdots a_1{a}_0}$, where $a_k$ is the quotient $q$. Note that $m$ has $k+1$ digits and $$\begin{array}{rl}m+2S(m)& =\overline{a_k{a}_{k-1}\cdots a_1{a}_0}+2(a_k+a_{k-1}+\cdots +a_1+a_0)=\\ & =10^k{a}_k+\overline{a_{k-1}\cdots a_1{a}_0}+2a_k+2(a_{k-1}+\cdots +a_1+a_0)=\\ & =(10^k+2)a_k+t+2S(t)=Nq+r=n,\end{array}$$ as desired.