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# 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

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

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

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.