Problem 10.5. Several sweet-tooths participated in a candy eating contest. Each participant ate an integer number of candies, and any two participants ate a different number of candies. Summarizing the contest, the jury ranked all the people in descending order of the number of candies eaten (for example, the winner ate the most candies, and the person who came in last place ate the fewest candies).

It is known that:

- the winner ate 14 times fewer candies than all the other participants combined;
- the participant who came in third place ate 20 times fewer candies than all the other participants combined;
- the participant who came in last place ate 21 times fewer candies than all the other participants combined.

How many sweet-tooths participated in the contest?
Answer: 21.

Solution. Let there be $n$ sweet-tooth participants in total, and they ate $S$ candies in total. From the condition, it follows that the winner ate $\frac{S}{15}$ candies, the person in third place $-\frac{S}{21}$ candies, and the person in last place $-\frac{S}{22}$ candies.

All participants, except the last one, ate more than $\frac{S}{22}$ candies. Therefore, $S > n \cdot \frac{S}{22}$, from which we get $n < 22$. On the other hand, since the person in third place ate $-\frac{S}{21}$ candies, and the person in last place ate $-\frac{S}{22}$ candies, we have $-\frac{S}{21} > -\frac{S}{22}$, which means $n \geqslant 21$.

Additionally, for $n=21$, we can provide an example. Suppose the 1st sweet-tooth participant ate 924 candies, the 2nd - 783 candies, the 3rd - 660 candies, and from the 4th to the 21st - from 647 to 630 candies respectively (a total of 13860 candies eaten). It is not difficult to verify that all conditions of the problem are satisfied.