# Sequence and fraction | AIME I, 2000 | Question 10

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

But try the problem first...

Answer: is 173.

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

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

\(100S-2(x_1+x_2+....+x_{100})=1+2+....+100\)

Second Hint

\(\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050\)

\(\Rightarrow S=\frac{2525}{49}\)

for \(k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}\)

Final Step

\(\Rightarrow x_{50}=\frac{75}{98}\)

\(\Rightarrow m+n\)=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