Final Round

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 $\frac{1}{14}\cdot (1+2+\cdots +13+100)=\frac{191}{14}=13\frac{9}{14}$. 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: $$\{1,2,3,4,5,6,7,8,9,10,11,12,18,100\}.$$ 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 $11$, etcetera. 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 $26$ etcetera, 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. $\square$