Given is an integer, what is the remainder when $$(6n + 3)^2$$ is divided by 9 ?
$$(6n + 3)^2=\left(3\left(2n+1\right)\right)^2=3^2.\left(2n+1\right)^2=9\left(2n+1\right)^2=9k$$
$$\therefore\ $$ $$(6n + 3)^2$$ is a perfect multiple of 9
$$=$$>Â The remainder when $$(6n + 3)^2$$ is divided by 9 is 0
Hence, the correct answer is Option D
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