4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar?
Answer: 16.

Solution. If the first jar contains no less than 11 beetles, then the second jar contains no less than 12, the third jar no less than 13, ..., and the tenth jar no less than 20. And in all jars, there are no less than $11+12+\ldots+20=155$ beetles. Contradiction. Therefore, the first jar contains no more than 10 beetles.

In the tenth jar, there are at least 9 more beetles than in the first jar, which means 9 beetles are no more than half the number of beetles in the tenth jar, so the first jar must contain no fewer than 9 beetles. If there are 9, then the tenth jar can only have 18. But $9+10+\ldots+18=135$. Contradiction. Therefore, the first jar contains exactly 10 beetles, and the tenth jar contains a maximum of 20. Since there are eleven numbers from 10 to 20 and $10+11+\ldots+20=165$, the variant with 15 beetles in a jar is missing, and the following are present: $10,11,12,13,14,16,17,18,19,20$. Therefore, the sixth jar contains 16 beetles.