Question 64

Given is an integer, what is the remainder when $$(6n + 3)^2$$ is divided by 9 ?

Solution

$$(6n + 3)^2=\left(3\left(2n+1\right)\right)^2=3^2.\left(2n+1\right)^2=9\left(2n+1\right)^2=9k$$

$$\therefore\ $$ $$(6n + 3)^2$$ is a perfect multiple of 9

$$=$$> The remainder when $$(6n + 3)^2$$ is divided by 9 is 0

Hence, the correct answer is Option D

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: