From the number 215, we can create a four-digit number by inserting any other digit between its digits. This way, we created two four-digit numbers, the sum of which was 4360. What could these two four-digit numbers be? Determine all possibilities.

(L. Simünek)
A newly created four-digit number is either of type $2 * 15$, or of type $21 * 5$. In the problem, the created numbers could be either a) both of the first type, or b) both of the second type, or c) each of a different type.

a) In this case, the sum $2 * 15 + 2 * 15$ ends with the number 30, which contradicts the given sum of 4360. Therefore, the numbers created in the problem could not both be of type $2 * 15$.

b) This case is solved as an alphametic:

$$
\begin{array}{cccc}
2 & 1 & * & 5 \\
2 & 1 & * & 5 \\
\hline 4 & 3 & 6 & 0
\end{array}
$$

In the hundreds column, we notice that after adding $1+1$, the digit 3 is recorded in the result, so there must have been a carry-over in the tens column. In the units column, we add $5+5=10$, and the digit 6 is recorded in the tens place in the result. Therefore, the places marked with asterisks must hide digits that sum to 15. This could be the sum of $9+6$ or $8+7$. Thus, this alphametic has two solutions.

c) This case is solved as an alphametic:

$$
\begin{array}{cccc}
2 & * & 1 & 5 \\
2 & 1 & * & 5 \\
\hline 4 & 3 & 6 & 0
\end{array}
$$

By similar reasoning as above, we determine that this alphametic has a single solution: the digit 4 belongs in the tens place, and the digit 2 belongs in the hundreds place.

The problem has a total of three solutions: the sought pair of numbers are either 2195 and 2165, or 2185 and 2175, or 2215 and 2145.

Scoring suggestion. 3 points for the solutions $2195 + 2165$ and $2185 + 2175$ (1 point for each solution and 1 point for the comment); 2 points for the solution $2215 + 2145$; 1 point for ruling out the possibility $2 * 15 + 2 * 15$.