If n is a natural number, then \[(6{{n}^{2}}+6n)\] is always divisible by |
A) Only 6
B) Both 6 and 12
C) Only 12
D) Only 18
Correct Answer: B
Solution :
\[(6{{n}^{2}}+6n)=6n\,(n+1)\] which is always divisible by 6 and 12 both, since \[n\,(n+1)\] is always even. |
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