Content

[hide]

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on positive solution.

## Positive solution – AIME I, 1990

Find the positive solution to

\(\frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}-\frac{2}{x^{2}-10x-69}=0\)

- is 107
- is 13
- is 840
- cannot be determined from the given information

**Key Concepts**

Integers

Divisibility

Algebra

## Check the Answer

But try the problem first…

Answer: is 13.

Source

Suggested Reading

AIME I, 1990, Question 4

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

here we put \(x^{2}-10x-29=p\)

\(\frac{1}{p}+\frac{1}{p-16}-\frac{2}{p-40}=0\)

Second Hint

or, (p-16)(p-40)+p(p-40)-2p(p-16)=0

or, -64p+(40)(16)=0

or, p=10

Final Step

or, 10=\(x^{2}-10x-29\)

or, (x-13)(x+3)=0

or, x=13 positive solution.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

Google