3. The last digit of a four-digit number was moved to the beginning (for example, $1234 \rightarrow$ 4123) and the resulting number was added to the original. The sum turned out to be 3333. What is the original number, given that its representation contains no digit 0? Find all possible options.
Answer: 1212 and 2121.

Solution. Let $\overline{x y t}$ be the desired four-digit number, and $\overline{t x y z}$ be the number obtained from it. Their sum is $S=\overline{x y z t}+\overline{t x y z}=11(100 x+10 y+z)+1001 t=11(\overline{x y z}+91 t)=3333$. From this, $\overline{x y z}+91 t=303$. The digit $t$ is different from zero and can only take the values 1 and 2. (If $t \geqslant 3$, then $\overline{x y z}<100$.) If $t=1$, then $\overline{x y z}=212$ and $\overline{x y z t}=2121$. If $t=2$, then $\overline{x y z}=121$ and $\overline{x y z t}=1212$.

Thus, the condition is satisfied by two numbers: 1212 and 2121.

Criteria. Finding at least one of the numbers - 2 points. Proving that the first (from the right) digit $t$ can only take the values 1 or 2 - another 2 points. Complete solution - 7 points.