Four different positive even numbers add up to 100. The third and fourth numbers are single-digit. If the first three numbers are multiplied by 12, 30, and 52 respectively, then the sum of these three resulting numbers is 2000. What are these four numbers?
Let the sought even numbers be: $2a, 2b, 2c, 2d$. Based on the problem, we can set up the following equations:

$$
\begin{aligned}
2a + 2b + 2c + 2d & = 100 \\
24a + 60b + 104c & = 2000
\end{aligned}
$$

In the second equation, every term can be divided by 4. Subtracting three times the first equation from the resulting equation, the sum of the terms containing $a$ will be 0:

$$
9b + 20c - 6d = 200
$$

From this,

$$
c = \frac{6d - 9b + 200}{20} = 10 - \frac{3}{20}(3b - 2d)
$$

For $c$ to be an integer, $(3b - 2d)$ must be divisible by 20. According to the problem, $2c$ is a positive single-digit integer, so $c$ can only be $1, 2, 3, 4$. Let's consider each case in turn.

1. If $c = 4$, then from the equation,

$$
3b - 2d = 40
$$

and thus,

$$
b = \frac{40 + 2d}{3} = 13 + \frac{1 + 2d}{3}
$$

Since $2d$ is also a single-digit number, $d$ can only be $1, 2, 3, 4$. According to the problem, the numbers must be different, so for $d$, only the values 1, 2, 3 are possible in this case. Among these, only $d = 1$ gives an integer value for $b$, which is 14. Substituting $b = 14, c = 4, d = 1$ into equation (1), we get 62 for $2a$, and thus the four numbers are:

$$
62, \quad 28, \quad 8, \quad 2
$$

2. If the value of $c$ is 3 or 2, it is easy to verify that in equation (3), $(3b - 2d)$ cannot be a multiple of 20 in either case. Therefore, we need to examine the case when
3. $c = 1$. Then $3b - 2d = 60$, and from this,

$$
b = \frac{60 + 2d}{3} = 20 + \frac{2d}{3}
$$

Since $d$ can only be 1, 2, 3, or 4, $b$ can only be an integer when $d = 3$, which gives 22. Calculating the value of $2a$ from equation (1), the four numbers in this case are:

$$
48, \quad 44, \quad 2, \quad 6
$$

Thus, our problem has two solutions, as it is easy to verify that both sets of four numbers satisfy the original conditions.

Németh Judit (Kecskemét, Közg. techn. III. o. t.)