Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.0 National Olympiad Prove it North Macedonia

Find the digits $A$ , $B$ , $C$ such that the given multiplication procedure is correct:
$\text{ABC BAC --A}$
Where $\overline{ABC} \cdot C = \overline{---A}$ , $\overline{ABC} \cdot A = \overline{---A}$ , $\overline{ABC} \cdot B = \overline{---B}$ . Distinct letters denote distinct digits and every "-" denotes one digit.

Solution

The numbers $ABC$ and $BAC$ are three-digit, so $A \neq 0$ and $B \neq 0$ . If $C=1$ the equality $\overline{ABC} \cdot C = \overline{---A}$ is not possible because $\overline{AB} \cdot 1 = \overline{AB} \neq \overline{AB} \cdot 0$ . If $A>3$ , from the equality $\overline{ABC} \cdot A = \overline{---A}$ it follows that $\overline{ABC} \cdot A > 400 \cdot 4 = 1600 > 999$ . Hence $A \leq 3$ . If $A=1$ , then $C=1$ which is impossible from the previous discussion. Therefore $A=2$ or $A=3$ . If $A=3$ then from $\overline{ABC} \cdot A = \overline{---A}$ it follows that $C=1$ which is impossible. Hence $A=2$ . Now, from $2\overline{BC} \cdot 2 = \overline{---2}$ , $C=1$ or $C=6$ . Therefore $C=6$ . Finally from the equality $2\overline{B6} \cdot B = \overline{---B}$ it must $B \in \{1,2,4,6,8\}$ . Checking all possibilities for $B$ gives that $B=8$ .

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