At a workplace, different capacity machines are working. On every possible pairing of two machines, they produce $35,39,40,49,44,46,30,41,32,36$ workpieces in one hour. How many workpieces do each of the machines produce in one hour?
I. Solution: If the number of machines is $x$, then according to the problem, $C_{x}^{2}=\frac{x(x-1)}{2}=10$, from which the only positive solution is $x=5$.

Therefore, we have 5 machines, which produce

$$
x_{1}<x_{2}<x_{3}<x_{4}<x_{5}
$$

workpieces per hour.

The number of workpieces produced by 2-2 machines in ascending order:

$$
30,32,35,36,39,40,41,44,46,49
$$

It is obvious that necessarily

$$
\begin{aligned}
& x_{1}+x_{2}=30 \\
& x_{1}+x_{3}=32 \\
& x_{4}+x_{5}=49 \\
& x_{3}+x_{5}=46
\end{aligned}
$$

Furthermore, it is obvious that

$$
\begin{gathered}
\left(x_{1}+x_{2}\right)+\left(x_{1}+x_{3}\right)+\left(x_{1}+x_{4}\right)+\left(x_{1}+x_{5}\right)+\left(x_{2}+x_{3}\right)+\left(x_{2}+x_{4}\right)+ \\
+\left(x_{2}+x_{5}\right)+\left(x_{3}+x_{4}\right)+\left(x_{3}+x_{5}\right)+\left(x_{4}+x_{5}\right)=392
\end{gathered}
$$

where 392 is the sum of the 10 numbers in (I).

Thus,

$$
4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=392
$$

or

$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=98
$$

From the system of five equations (1), (2), (3), (4), (5), we get:

$$
x_{1}=13, \quad x_{2}=17, \quad x_{3}=19, \quad x_{4}=22, \quad x_{5}=27
$$

By verification, we can confirm that these values indeed satisfy the conditions of the problem.

István Kovács (Kecskemét, Ságvári specialized college)

II. Solution: In the sequence (1) of ten numbers representing the sum of the performance of 2-2 machines, we find 4 odd and 6 even numbers, which means that out of the five machines' performances, 1 is even (or odd) and 4 are odd (or even). Any other assumption regarding parity contradicts the property of the sequence (1) mentioned above, which can be verified with a brief consideration.

Let the performance of the 5 machines be denoted by $x, y, z, u$, and $v$, and let $x$ be the one with a different parity from the other 4, then

$$
\begin{array}{lcl}
x+y=35 & \text { from which } & y=35-x \\
x+z=39 & » & z=39-x \\
x+u=41 & » & u=41-x \\
x+v=49 & » & v=49-x
\end{array}
$$

from which
»
»
»

The sum of the other 6 combinations

$$
\begin{gathered}
(y+z)+(y+u)+(y+v)+(z+u)+(z+v)+(u+v)=30+32+36+ \\
+40+44+46=228
\end{gathered}
$$

thus

$$
3(y+z+u+v)=228
$$

from which

$$
y+z+u+v=76
$$

If we substitute the values of (1), (2), (3), (4) into (5), we get

$$
164-4 x=76
$$

from which

$$
x=22
$$

and thus

$$
y=35-22=13, \quad z=39-22=17, \quad u=41-22=19, \quad \text { and } \quad v=49-22=27
$$