## Task 1 - 160821

For student experiments, exactly 29 individual parts (experimental materials) were purchased for exactly $29 \mathrm{M}$. These were parts costing 10 M, 3 M, or 0.50 M; at least one part of each type was included. No other types were among the purchased items.

How many parts of each of the three types were there in total?
If 3 or more parts cost 10 M more than 29 M, then there were at most 2 parts for $10 \mathrm{M}$.

Assuming there were exactly 2 parts for $10 \mathrm{M}$. Then exactly $9 \mathrm{M}$ would remain for the two remaining types. From these, 1 or 2 parts for $3 \mathrm{M}$ could have been purchased, after which $6 \mathrm{M}$ or $3 \mathrm{M}$ would remain for the parts costing $0.50 \mathrm{M}$. This would be 12 or 6 parts for $0.50 \mathrm{M}$, and thus a total of 15 or 10 individual parts, which is too few.

Therefore, exactly 1 part for $10 \mathrm{M}$ was purchased, and exactly $19 \mathrm{M}$ remained for the $3 \mathrm{M}$ parts and $0.50 \mathrm{M}$ parts.

Assuming only 1 part for $3 \mathrm{M}$ was purchased, then $16 \mathrm{M}$ would remain, for which 32 parts for $0.50 \mathrm{M}$ could be purchased, thus a total of 34 parts, in contradiction to the problem.

If the number of parts for $3 \mathrm{M}$ is increased by 1 each time, then, to keep the total price the same, the number of parts for $0.50 \mathrm{M}$ decreases by 6 each time, and the total number of parts decreases by exactly 5. The only way to reach 29 parts in this manner is to increase the number of parts for $3 \mathrm{M}$ by exactly 1, and thus decrease the number of parts for $0.50 \mathrm{M}$ by exactly 6. Therefore, exactly 1 part for $10 \mathrm{M}$, exactly 2 parts for $3 \mathrm{M}$, and exactly 26 parts for $0.50 \mathrm{M}$ were purchased.