Problem 3. In a five-digit number, each digit was increased by 2 or by 4 (different digits could be increased by different numbers), as a result of which the number increased fourfold. What could the original number have been? Find all possible options and prove that there are no others.
Solution. Answer: 14074.

Let the original five-digit number be denoted by $N$, then the resulting number is equal to $4N$. Their difference is $3N$ and is a five-digit number consisting of 2 and 4. This difference $3N$ is not less than $3 \cdot 10000$, so it starts with a four.

By the divisibility rule for 3, the sum of the digits of the number $3N$ is divisible by 3, so it cannot consist of two fours and three twos, three fours and two twos, or five fours. Therefore, the number $3N$ consists either of one four and four twos, or of four fours and one two. Then the number $N$ can be one of the numbers $\frac{42222}{3}=14074, \frac{44442}{3}=14814, \frac{44424}{3}=14808, \frac{44244}{3}=14748$, $\frac{42444}{3}=14148$.

It is also clear that the number $N$ could not contain an eight in its decimal representation, so only the number 14074 remains to be checked. And it fits: $14074 \cdot 4 = 56296$, all the digits of the quadrupled number are greater than the corresponding original digits by 2 or 4.

## Criteria

The highest applicable criterion is used:

7 points. Any complete solution of the problem.

6 points. A generally correct solution, but an arithmetic error is made that does not affect its course.

5 points. The numbers $14074, 14814, 14808, 14748, 14148$ are obtained justifiably.

4 points. It is proven that the number is increased by either one four and four twos, or four fours and one two.

1 point. There is only the answer.