Question:

Neha lists all positive divisors of $(2010)^2$. She randomly selects 2 distinct divisors. Probability that exactly one is a perfect square is

$\dfrac{32}{81}$

$\dfrac{26}{81}$

$\dfrac{16}{81}$

$\dfrac{34}{81}$

Step-by-Step Solution

Key Concept: $(2010)^2=2^2\cdot3^2\cdot5^2\cdot67^2$. Total divisors $=(2+1)^4=81$. Perfect square divisors$=(1+1)^4=16$. Non-square$=65$.

$P=\dfrac{26}{81}$.

Correct Answer: 2