2. We call a collection of at least two different positive integers eternal if the largest number is equal to 100. We will look at the average of all numbers in an eternal collection, which we call the average of the collection. For example: the average of the eternal collection $\{1,2,20,100\}$ is $\frac{123}{4}$ and the average of the eternal collection $\{74,90,100\}$ is 88.

Determine all integers that can occur as the average of an eternal collection.
2. We will solve these problems in two parts. First, we will show that the smallest possible integer average of an eternal set is 14, and then we will demonstrate that you can obtain all integers greater than or equal to 14 but less than 100 as the average of an eternal set.

If you decrease one of the numbers (not equal to 100) in an eternal set, the average decreases. Also, if you add a number to an eternal set and that number is smaller than the current average, the average will decrease. To find the eternal set with the smallest average, we can start with 1,100 and then keep adding the smallest possible numbers until the next number we would add is greater than the current average. We find that the set with 1 through 13 and 100 has an average of \(\frac{1}{14} \cdot (1+2+\ldots+13+100) = \frac{191}{14} = 13 \frac{9}{14}\). Adding 14 would increase the average, and removing 13 (or more numbers) would also increase the average. We conclude that the average of an eternal set must be at least 14 if it is to be an integer.

The smallest integer that can be the average of an eternal set is therefore 14, with (for example) the following eternal set:

\[
\{1,2,3,4,5,6,7,8,9,10,11,12,18,100\} \text{.}
\]

What we now need to show is that all integers greater than 14 (and less than 100) can also be the average of an eternal set. We start with the eternal set above with an average of 14. Each time you add 14 to one of the numbers in this eternal set, the average increases by 1. Do this from right to left, starting by adding 14 to 18 (the average then becomes 15), then adding 14 to 12 (the average then becomes 16), then adding 14 to 11, and so on. Eventually, you will get the eternal set \(\{15,16,17,18,19,20,21,22,23,24,25,26,32,100\}\) with an average of 27, and you will have encountered all values from 14 to 27 as averages. Because we started with the highest number after 100, the sequence of numbers remains increasing, ensuring that it always consists of 14 different numbers and thus forms a valid eternal set.

We can do this again by first adding 14 to 32, then adding 14 to 26, and so on, and we will end up with a set with an average of 40. If we do this one more time, we will eventually get the set \(\{43,44,45,46,47,48,49,50,51,52,53,54,60,100\}\) with an average of 53. We can also get 54 as the average of the eternal set \(\{8,100\}\), 55 as the average of \(\{10,100\}\), and so on up to 99 as the average of \(\{98,100\}\). With this, we have shown that we can obtain all values from 14 to 99.