Find two consecutive multiples of 3 whose product is 270.

Given,

Let two consecutive multiples of 3 be 3n and 3n + 3.

Given,

The product is 270.

⇒ 3n(3n + 3) = 270

⇒ n(3n + 3) = $\dfrac{270}{3}$

⇒ 3n² + 3n = 90

⇒ 3n² + 3n - 90 = 0

⇒ 3(n² + n - 30) = 0

⇒ n² + n - 30 = 0

⇒ n² + 6n - 5n - 30 = 0

⇒ n(n + 6) - 5(n + 6) = 0

⇒ (n - 5)(n + 6) = 0

⇒ (n - 5) = 0 or (n + 6) = 0     [Using zero-product rule]

⇒ n = 5 or n = -6

Case 1: n = 5

First multiple is 3n = 3(5) = 15

Second multiple is 3n + 3 = 3(5) + 3 = 15 + 3 = 18

Case 2: n = -6

First multiple is 3n = 3(-6) = -18

Second multiple is 3n + 3 = 3(-6) + 3 = -18 + 3 = -15

Hence, two consecutive multiples of 3 are 15 and 18 or -18 and -15.