If $n$ is a natural number, then 6n2 + 6n is always divisible by:

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If $n$ is a natural number, then 6n² + 6n is always divisible by:

6 only

6 and 12 both

12 only

by 18 only

Explanation:

6$n$² + 6$n$ = 6$n$ $\left(n + 1\right)$, which is always divisible by 6 and 12 both, since $n$ $\left( n + 1\right)$ is always even.

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On dividing a number by 357, we get 39 as remainder. On dividing the same number 17, what will be the remainder ?

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