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