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.
Problem
Source: Dutch NMO 2022 p2
Tags: number theory
18.11.2022 01:08
The minimal sum for $n$ numbers in the list, is $\frac{100 + \frac{n(n+1)}{2}}{n+1} = \frac{200+n^2+n}{2n+2}$, which is minimized for $n = 10\sqrt{2}-1$, giving us $10\sqrt{2}-1$ in return. This means that every number $\ge 14$ could be reached in theory. Now the upperbound: The maximal sum clearly is $99$. In theory, we can reach every number between $13$ and $100$, we just need to find configurations. This is pretty trivial to be honest. If the sum is too small, you can replace with larger one etc.
18.11.2022 01:31
As configuration we can use next sets If $14 \leq n\leq 50$ then set $\{n-13,n-12,n-11,...,n-2,2n-10,100 \}$ For $51 \leq n \leq 99$ we can use set $\{2n-100,100\}$
28.08.2023 00:54
If the maximum is 100 then the average cannot be greater than 100 but since the numbers are never said to be different then we can have a set of several 100 and their average would be 100. The larger the set, the smaller the average can be. For example, if we have a set of 100 elements, then we can divide any multiple of 100 into 99 and we will then have the numbers from 2 to 100. like this: The average of $\{1,1,1,1,...,2,100\}$ is 2 The average of $\{2,2,2,2,...,4,100\}$ is 3 we cannot make the average 1 because there will always be 99 left over For example, in a set of 1000 we have that 999 ones plus 100 is equal to 1099 and for the average it is divided by 1000