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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.

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

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

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