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# Probability of divisors | AIME I, 2010 | Question 1

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

Series

Probability

Number Theory

AIME I, 2010, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

$2010^{2}=2^{2}3^{2}5^{2}67^{2}$

Second Hint

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

Final Step

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