If $n$ is a natural number, then $(6 n^{2} + 6 n)$ is always divisible by:

6

only

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6

and

12

both

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12

only

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18

only

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Solution

The correct option is B $6$ and $12$ both
$(6 n^{2} + 6 n) = 6 n (n + 1)$ , which is always divisible by $6$ and $12$ both, since $n (n + 1)$ is always even.

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Similar questions

A number is divisible by both $5$ and $12$ . By which other number will that number be always divisible?

n

n (n + 1) (n + 2)

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