8.2. In a five-digit number, one digit was crossed out, and the resulting four-digit number was added to the original. The sum turned out to be 54321. Find the original number.
Answer: 49383. Solution. Note that the crossed-out digit must be the last digit of the number $N$, because otherwise the sum of the two numbers would have an even digit in the last place. Let's denote this crossed-out digit by $x$ and let $y$ be the four-digit number obtained after the crossing out. Then the condition can be written as: $10 y + x + y = 54321 \Leftrightarrow 11 y + x = 54321$. Dividing 54321 by 11 with a remainder, we get $54321 = 11 \cdot 4938 + 3$ and considering that $x < 11$, we will have the unique solution (due to the uniqueness of division with a remainder) $x = 3, y = 4938$.