4. How many three-digit numbers exist that are 5 times the product of their digits?
Answer. One number is 175.

Solution. First method. The digits that make up the number do not include the digit 0, otherwise the condition of the problem cannot be met. The given three-digit number is obtained by multiplying 5 by the product of its digits, so it is divisible by 5. Therefore, its notation ends with the digit 5. We get that the product of the digits, multiplied by 5, must be divisible by 25. Note that there cannot be any even digits in the notation of the number, otherwise the product of the digits would be zero. Thus, the three-digit number must be divisible by 25 and not contain even digits. There are only five such numbers: $175, 375, 575, 775$, and 975. The product of the digits of the desired number must be less than 200, otherwise, when multiplied by 5, it would give a four-digit number. Therefore, the numbers 775 and 975 are definitely not suitable. Among the remaining three numbers, only 175 satisfies the condition of the problem.

Second method. Note (similar to the first method of solving) that the last digit of the desired number is 5. Let $a, b, 5$ be the consecutive digits of the desired number. According to the problem, we have: $100a + 10b + 5 = a \cdot b \cdot 5 \cdot 5$. Dividing both sides of the equation by 5, we get: $20a + 2b + 1 = 5ab$. After subtracting $20a$ from both sides of the equation and factoring out the common factor on the right side, we get: $2b + 1 = 5a(b - 4a)$ (1). Considering that $a$ and $b$ can take natural values from 1 to 9, we get that the possible values of $a$ are only 1 or 2. But $a = 2$ does not satisfy equation (1), as the left side is an odd number, and the right side, when substituting $a = 2$, is even. Therefore, the only possibility is $a = 1$. Substituting this value into (1), we get: $2b + 1 = 5b - 20$, from which $b = 7$. Answer: the only desired number is 175.

## Grading Criteria.

- Complete correct solution - 7 points.
- Correct answer obtained and there are arguments that significantly reduce the number of cases to check, but the full solution is not provided - 4 points.
- The equation is correctly set up and transformations and arguments are provided that allow solving the problem, but the solution is not completed - 4 points.

All-Russian School Olympiad 2014-2015.

- The number of cases to check is reduced, but there are no explanations why, and the correct answer is given - 3 points.
- The equation is correctly set up, but the problem is not solved - 2 points.
- There are arguments in the solution that allow excluding certain numbers or considering numbers with certain properties (for example, ending in the digit 5), but there is no further significant progress in solving the problem - 1 point.
- Only the correct answer or the answer with verification is provided - 1 point.