The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers.
Let the two consecutive natural numbers, which are multiples of 3, be $x$ and $x+3$
Then, as per question
$x \left(x+3\right) = 810$
i.e. $x^2 + 3x - 810 = 0$
i.e. $x^2 -27x + 30x -810 = 0$
i.e. $x \left(x-27\right)+30\left(x-27\right)=0$
i.e. $\left(x-27\right)\left(x+30\right)=0$
$\implies$ $x=27$ or $x=-30$
Since the required numbers are natural numbers, therefore, $x=27$ is the acceptable solution.
$\therefore$ The required numbers are 27, 30