Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.

## Probability of divisors – AIME I, 2010

Ramesh lists all the positive divisors of \(2010^{2}\), she then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers. Find m+n.

- is 107
- is 250
- is 840
- cannot be determined from the given information

**Key Concepts**

**S**eries

Probability

Number Theory

## Check the Answer

Answer: is 107.

AIME I, 2010, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

\(2010^{2}=2^{2}3^{2}5^{2}67^{2}\)

\((2+1)^{4}\) divisors, \(2^{4}\) are squares

probability is \(\frac{2.2^{4}.(3^{4}-2^{4})}{3^{4}(3^{4}-1)}=\frac{26}{81}\) implies m+n=107

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s