The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers.

Given,

Let two consecutive natural numbers multiple of 3 be 3n and 3n + 3.

Given,

The product is 810.

⇒ 3n(3n + 3) = 810

⇒ n(3n + 3) = $\dfrac{810}{3}$

⇒ 3n² + 3n = 270

⇒ 3n² + 3n - 270 = 0

⇒ 3(n² + n - 90) = 0

⇒ n² + n - 90 = 0

⇒ n² + 10n - 9n - 90 = 0

⇒ n(n + 10) - 9(n + 10) = 0

⇒ (n - 9)(n + 10) = 0

⇒ (n - 9) = 0 or (n + 10) = 0     [Using zero-product rule]

⇒ n = 9 or n = -10.

n = 9 [Since the numbers are natural numbers]

First multiple is 3n = 3(9) = 27

Second multiple is 3n + 3 = 3(9) + 3 = 27 + 3 = 30.

Hence, two consecutive natural multiples of 3 are 27 and 30.