[ Inequality problems. Case analysis]

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?

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Let there be $x$ bugs in the first jar, then in the second jar - no less than $x+1$ bug, in the third jar - no less than $x+2$ bugs, ..., in the tenth jar no less than $x+9$ bugs. Therefore, the total number of bugs is no less than $10 x+45$. Considering that a total of 150 bugs were distributed, we get $x \leq 10$.

On the other hand, in the tenth jar there should be no more than $2 x$ bugs, in the ninth - no more than $2 x-1$ bugs, and so on. This means that in the first jar - no more than $2 x-9$ bugs, and the total number of bugs - no more than $20 x-45$. Since a total of 150 bugs were distributed, then $x \geq 10$.

Thus, there are exactly 10 bugs in the first jar, and 19 or 20 in the last jar. Let's find the sum of eleven consecutive numbers starting from ten: $10+11+\ldots+19+20=165$. Since there should be a total of 150 bugs, the jar with 15 bugs is missing. Therefore, the distribution is uniquely determined: 10, 11, 12, 13, 14, 16, 17, 18, 19, and 20 bugs from the first to the tenth jar respectively. Thus, there are 16 bugs in the sixth jar.

## Answer

16 bugs.

## Problem