36th Hellenic Mathematical Olympiad

Let $\kappa$ be the number of digits of the integer $A$ which is equal to $13$ times the sum of its digits. The least possible $A$ is $10^{\kappa -1}$, while the maximal possible sum of the digits is $9\kappa$. Therefore we need to have: $$10^{\kappa -1}\le 13\cdot 9\kappa =117\kappa .(1)$$ For $\kappa \ge 4$, we will prove using induction that: $10^{\kappa -1}>117\kappa$, that is, relation (1) is not valid. In fact, for $\kappa =4$ we have: $10^{4-1}=10^3>117\cdot 4=468$. If $10^{\kappa -1}>117\kappa$, for the arbitrary $\kappa >4$, then we get: $$10^{(\kappa +1)-1}=10^{\kappa }=10\cdot 10^{\kappa -1}>10\cdot 117\kappa =117\cdot 10\kappa >117\cdot (\kappa +1).$$ Therefore the number $\kappa$ must be less or equal to $3$.

We have to reject the case $\kappa =1$, since $A=\alpha <13\alpha$, with $0<\alpha \le 9$. Similarly we reject the case with $\kappa =2$, since $A=10\alpha +\beta <13(\alpha +\beta )$. Let $\kappa =3$ and $A=\overline{\alpha \beta \gamma }=100\alpha +10\beta +\gamma$, $0<\alpha \le 9$, $0\le \beta ,\gamma \le 9$.

Then we have: $$100\alpha +10\beta +\gamma =13\cdot (\alpha +\beta +\gamma ),0<\alpha \le 9,0\le \beta ,\gamma \le 9$$ $$\iff 87\alpha =3\beta +12\gamma ,0<\alpha \le 9,0\le \beta ,\gamma \le 9$$ $$\iff 29\alpha =\beta +4\gamma ,0<\alpha \le 9,0\le \beta ,\gamma \le 9$$ Since $0\le \beta +4\gamma \le 45\Rightarrow 29\alpha \le 45\Rightarrow \alpha \le 1$ (since $\alpha \ne 0$), and hence: $$\beta +4\gamma =29\Rightarrow \beta =29-4\gamma \ge 0\Rightarrow 0\le \beta =29-4\gamma \le 9$$ $$\Rightarrow -29\le -4\gamma \le -20\Rightarrow 5\le \gamma \le \frac{29}{4}\Rightarrow \gamma \in \{5,6,7\}$$ For $\gamma =5\Rightarrow \beta =9$ and $A=195$. For $\gamma =6\Rightarrow \beta =5$ and $A=156$. For $\gamma =7\Rightarrow \beta =1$ and $A=117$.

Therefore, the positive integers which are equal to $13$ times the sum of their digits are $117$, $156$, and $195$.