At a workplace, different performance automated machines are in operation. In all possible pairs of these machines, they produce $35,39,40,49,44,46,30,41,32,36$ workpieces in one hour. How many machines are working in total, and how many workpieces does each machine produce in one hour?
a) If every machine operator writes down the combined performance of their machine with every other machine, then on the one hand, every data point will appear in two records, so all the records together will contain 20 data points. On the other hand, there will be as many records as there are machines, and each record will contain one less data point than the number of machines. Therefore, 20 must be factored into a number and a number one less than it. The only such factorization is $20=5 \cdot 4$, because if one factor is increased, the other must be decreased to keep the product constant. Thus, there are 5 machines.

b) According to this, the sum of the ten paired results gives each machine's 4-hour production, so the sum gives the 4-hour production of the 5 machines. The sum is 392, so the combined 1-hour production of the 5 machines is 98 units.

There are no two machines with the same performance, otherwise, pairing two such with a third would result in the same performance, but all the data points are different. Let's arrange the 5 machines in order of increasing performance.

The smallest data point is 30, which is clearly the combined performance of the first (slowest) machine with the second. Similarly, the largest result, 49, is the combined performance of the V. and IV. machines. Subtracting the sum of these two pairs from the combined performance of the 5 machines gives the performance of the III. machine: $98-(30+49)=19$.

The smallest data point after 30 is 32 units, which can only be the combined production of the I. and III. machines (otherwise, their combined production would be less, but there is no smaller data point). Thus, the I. machine produces $32-19=13$ units per hour, and the II. machine produces $30-13=17$ units.

Similarly, the largest data point after 49, 46 units, can only result from pairing the V. and III. machines, so the V. machine produces $46-19=27$ units, and the IV. machine produces $49-27=22$ units. Therefore, the performance of the machines in order is

$13,17,19,22,27$

These can form the sums of the pairs not yet mentioned, i.e., I-IV, I-V, II-III, II-IV, II-V, III-IV, which are exactly the other results in the data series, so our solution is correct.

Fencsik Gábor (Budapest, Berzsenyi D. g. I. o. t.)

Note. We only used the two smallest, the two largest data points, and the sum of the data, and as the solution process shows, these already determine the other data points. Our data, therefore, are not independent. If we changed some of the data points but in such a way that the above 5 values did not change, the problem would no longer have a solution.