Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.
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.
Series
Probability
Number Theory
Answer: is 107.
AIME I, 2010, Question 1
Elementary Number Theory by Sierpinsky
\(2010^{2}=2^{2}3^{2}5^{2}67^{2}\)
\((2+1)^{4}\) 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
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