Problem:
A finite set S of distinct real numbers has the following properties: the mean of S∪{1} is 13 less than the mean of S, and the mean of S∪{2001} is 27 more than the mean of S.
Solution:
Let S have n elements with mean x. Then
n+1nx+1=x−13 and n+1nx+2001=x+27
or
nx+1=(n+1)x−13(n+1) and nx+2001=(n+1)x+27(n+1)
Subtract the third equation from the fourth to obtain 2000=40(n+1), from which n=49 follows. Thus x=651.
The problems on this page are the property of the MAA's American Mathematics Competitions