
2. A set consisting of at least two distinct positive integers is called centenary if its greatest element is 100 . We will consider the average of all numbers in a centenary set, which we will call the average of the set. For example, the average of the centenary set $\{1,2,20,100\}$ is $\frac{123}{4}$ and the average of the centenary set $\{74,90,100\}$ is 88 .

Determine all integers that can occur as the average of a centenary set.


2. We solve this problem in two steps. First we will show that the smallest possible integral average of a centenary set is 14 , and then we will show that we can obtain all integers greater than or equal to 14 , but smaller than 100 , as the average of a centenary set.

If you decrease one of the numbers (unequal to 100) in a centenary set, the average becomes smaller. Also if you add a number that is smaller than the current average, the average becomes smaller. To find the centenary set with the smallest possible average, we can start with 1,100 and keep adjoining numbers that are as small as possible, until the next number that we would add is greater than the current average. In this way, we find the set with the numbers 1 to 13 and 100 with average $\frac{1}{14} \cdot(1+2+\ldots+13+100)=\frac{191}{14}=13 \frac{9}{14}$. Adding 14 would increase the average, and removing 13 (or more numbers) would increase the average as well. We conclude that the average of a centenary set must be at least 14 when it is required to be an integer.

Therefore, the smallest integer which could be the average of a centenary set is 14 , which could for example be realised using the following centenary set:

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

Now we still have to show that all integers greater than 14 (and smaller than 100) can indeed be the average of a centenary set. We start with the centenary set above with average 14. Each time you add 14 to one of the numbers in this centenary set, the average increases by 1. Apply this addition from right to left, first adding 14 to 18 (the average becoming 15), then adding 14 to 12 (the average becoming 16), then adding 14 to 11, et cetera. Then you end up with the centenary set $\{15,16,17,18,19,20,21,22,23,24,25,26,32,100\}$ with average 27 , and you realised all values from 14 to 27 as an average. Because we started adding 14 to the second largest number in the set, this sequence of numbers remains increasing during the whole process, and therefore consists of 14 distinct numbers the whole time, and hence the numbers indeed form a centenary set.

We can continue this process by first adding 14 to 32 , then 14 to 26 et cetera, and then we get a centenary set whose average is 40 . Repeating this one more time, we finally end up with the set
$\{43,44,45,46,47,48,49,50,51,52,53,54,60,100\}$ with average 53. Moreover, we can obtain 54 as the average of the centenary set $\{8,100\}, 55$ as the average of $\{10,100\}$, and so on until 99 , which we obtain as the average of $\{98,100\}$. This shows that all values from 14 to 99 can be obtained.
