XXVII Olimpiada Matemática Rioplatense

Let $N$ be the number on the blackboard. Then, the number written by Carlos in his notebook is $C=4N+30$. Call $D$ the number written by Dora. Since $D=C\ge N$ is a four digit number, the same holds for $C$; then, the first digit of $N$ is not greater than $2$, since, otherwise, $C\ge 4\cdot 3000=12000$. On the other hand, it is clear that $4N+30$ is even. So, the last digit of $D$, which is the first digit of $N$, is even (and greater than $0$). Combining both previous remarks, we conclude that the first digit of $N$ is $2$. Since $D\ge 4N\ge 4\cdot 2000=8000$, we also deduce that the first digit of $D$ is at least $8$. As adding $30$ does not change the last digit of a number, we have that $D$ and $4N$ have the same last digit, which is $2$. In order for $4N$ to end with $2$, $N$ has to end with $2$ or $8$. But the last digit of $N$ is the first digit of $D$, which we know is at least $8$. Then, the only possibility is that $N$ ends with $8$. Thus, the number on the blackboard is of the form $N=\overline{2ab8}=2000+100a+10b+8$, whereas $D=\overline{8ba2}=8000+100b+10a+2$. Now, the relation $D=C$ can be restated as $$8000+100b+10a+2=8000+400a+40b+32+30,$$ which simplifies to $60b-60=390a$, and dividing by $30$, we obtain $$2b-2=13a.$$ Since $b$ is a digit, we have $-2\le 2b-2\le 2\cdot 9-2=16$. The only multiples of $13$ in this interval are $0$ and $13$, but taking into account that $2b-2$ is even, we conclude that the only possibility is that $2b-2=0$. Then, $b=1$ and $a=0$. Therefore, $N=2018$, which satisfies the condition and is the unique solution.