Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Number Theory and Probability.

Number Theory and Probability – AIME 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


But try the problem first…

Answer: is 107.

Source
Suggested Reading

AIME, 2010, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

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

Second Hint

There are \((2+1)^{4}\) divisors, \(2^{4}\)of which 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

Subscribe to Cheenta at Youtube