Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.
A sequence of numbers \(x_1,x_2,....,x_{100}\) has the property that, for every integer k between 1 and 100, inclusive, the number \(x_k\) is k less than the sum of the other 99 numbers, given that \(x_{50}=\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.
Equation
Algebra
Integers
Answer: is 173.
AIME I, 2000, Question 10
Elementary Number Theory by Sierpinsky
Let S be the sum of the sequence \(x_k\)
given that \(x_k=S-x_k-k\) for any k
taking k=1,2,....,100 and adding
\(100S-2(x_1+x_2+....+x_{100})=1+2+....+100\)
\(\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050\)
\(\Rightarrow S=\frac{2525}{49}\)
for \(k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}\)
\(\Rightarrow x_{50}=\frac{75}{98}\)
\(\Rightarrow m+n\)=75+98
=173.
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