Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.
Sequence and fraction – AIME I, 2000
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.
- is 107
- is 173
- is 840
- cannot be determined from the given information
Key Concepts
Equation
Algebra
Integers
Check the Answer
Answer: is 173.
AIME I, 2000, Question 10
Elementary Number Theory by Sierpinsky
Try with Hints
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.
Other useful links
- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s