Problem:
A set S of distinct positive integers has the following property: for every integer x in S, the arithmetic mean of the set of values obtained by deleting x from S is an integer. Given that 1 belongs to S and that 2002 is the largest element of S, what is the greatest number of elements that S can have?
Solution:
Let x1​,x2​,x3​,…,xn​ be the members of S, and let
sj​=n−1x1​+x2​+x3​+⋯+xn​−xj​​
It is given that sj​ is an integer for any integer j between 1 and n, inclusive. Note that, for any integers i and j between 1 and n, inclusive,
si​−sj​=n−1xj​−xi​​
which must be an integer. Also, xj​=(si​−sj​)(n−1)+xi​, and when xi​=1, this implies that each element of S is 1 more than a multiple of n−1. It follows that (n−1)2+1≤2002, implying that n≤45. Since n−1 is a divisor of 2002−1, conclude that n=2 or n=4 or n=24 or n=30, so n is at most 30​. A thirty-element set S with the requested property is obtained by setting xj​=29j−28 for 1≤j≤29 and x30​=2002.
The problems on this page are the property of the MAA's American Mathematics Competitions