Problem:
A sequence of numbers x1,x2,x3,…,x100 has the property that, for every integer k between 1 and 100, inclusive, the number xk is k less than the sum of the other 99 numbers. Given that x50=m/n, where m and n are relatively prime positive integers, find m+n.
Solution:
Let S=x1+x2+x3+⋯+x100. so xk=(S−xk)−k for all integers k between 1 and 100. inclusive. Thus k+2xk=S for all such k. Summing these equations for k=1,2,3, ... 100 yields
2100⋅101+2S=100S
from which S=492525 follows. Thus x50=2S−50=9875, and m+n=173.
The problems on this page are the property of the MAA's American Mathematics Competitions