Task 7.2. Solve the puzzle

$$
A B C D E F \cdot 3 = B C D E F A
$$

List all the options and prove that there are no others.
Answer: $142857 \cdot 3=428571$ and $285714 \cdot 3=857142$.

Solution. Let the five-digit number $BCDEF$ be denoted by $x$. Then this puzzle can be rewritten as the relation $(100000 \cdot A + x) \cdot 3 = 10x + A$, which is equivalent to the equation $29999 \cdot A = 7x$, or $42857 \cdot A = x$.

Notice that digits $A \geqslant 3$ do not work, because the number $42857 \cdot A$ would be a six-digit number, which $x$ cannot be.

For $A=1$, we get that $42857 = BCDEF$, i.e., $B=4, C=2, D=8, E=5, F=7$.

For $A=2$, we get that $2 \cdot 42857 = 85714 = BCDEF$, i.e., $B=8, C=5, D=7, E=1, F=4$.

Both of these solutions are valid.

## Criteria

The highest applicable criterion is used:

7 6. Any correct solution to the problem.

3 6. An equation of the form $(100000 \cdot A + x) \cdot 3 = 10x + A$ or equivalent is set up.

5 6. Such an equation is set up, and one of the solutions is correctly found.

## Kurchatov Olympiad in Mathematics - 2019 Final Stage.

In the absence of the above advancements, the following criteria are summed:

## 16 . The correct answer is provided.

2 6. It is proven that $A=1$ or $A=2$.

If only proven that $A \leqslant 3$, then 1 point is given instead of 2.