Question

Maya 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 $m + n.$

答案

Solution:

Since $2010^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 67^2,$ it has $(2+1)^4 = 81$ positive divisors. A divisor is a perfect square exactly when each of its four exponents is $0$ or $2,$ giving $2^4 = 16$ perfect squares and $81 - 16 = 65$ non-squares.

The probability of picking one of each is $p = \frac{16 \cdot 65}{\binom{81}{2}} = \frac{1040}{3240} = \frac{26}{81},$ so $m + n = 26 + 81 = 107.$

2010 AIME I Problem 1

Solution:

Problem 1 in Other Years