7.3 In a six-digit number, one digit was crossed out to obtain a five-digit number. The five-digit number was subtracted from the original number, and the result was 654321. Find the original number
Answer: 727 023. Solution: Note that the last digit was crossed out, as otherwise the last digit of the number after subtraction would have been zero. Let $y$ be the last digit of the original number, and $x$ be the five-digit number after crossing out. Then the obtained number is $10 x + y$ $x = 9 x + y = 654321$. Dividing this number by 9 with a remainder (and considering that $y$ does not exceed 9), we get the remainder $y = 3$ and the quotient $x = 72702$.