Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Romania

Let $n \ge 2$ be an integer. We define the numbers $A = 33...3$ , with $n$ digits $3$ , and $B = 20 \cdot A + 6$ . Find all the digits that form the number $A \cdot B$ .

Solution

Notice that $3 \cdot A = 99...9 = 10^n - 1$ . Then $A \cdot B = (10^n - 1) \cdot \underbrace{22...2}_{n+1 \text{ digits}} = \underbrace{22...2}_{n+1 \text{ digits}} \underbrace{00...0}_{n \text{ digits}} - \underbrace{22...2}_{n+1 \text{ digits}}$ . Thus $A \cdot B = \underbrace{22...2}_{n-1 \text{ digits}} \underbrace{1977...78}_{n-1 \text{ digits}}$ .
So, the digits of $A \cdot B$ are $1$ , $2$ , $7$ , $8$ and $9$ .

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