Aptitude - Numbers - Discussion

Hey I don't understand this problem please help!

I don't understand this problem please help.

If a number is divisible by 12 or 18 then it is always divisible by 6.
So option C and D are eliminated...

We can see the no. 6n(n+1) is divisible by 6 so we have to check for 12
Now a number would be divisible by 12 if it is multiplication of 6 and a even number. (e.g. (6*even number)

Now n(n+1) is even number.
So answer is B

How is n(n+1) proved an even numer?

Product of even, odd is always even.

ELIMINATION METHOD...

PUT n=1. we get 12.
12 is divisible by both 12, 6.

But, If we put n=9 in 6n2+6n.

i.e, 6(9)2+6(9)

= 6(81)+54

= 486+54

= 540

But, here 540 is not divisible by 12.

So, how?

Who told you that 540 is not divisible by 12?

540/12 = 45.

Firstly, we have to use 6n(n+1).

Where as n= natural number, we can use any number what you desire,
For eg 1, when n=2, 6*2(2+1) =12*3 =36,

36 can be divided by 6 , 12 and 18.

Eg 2, where n=1, 6*1(1+1)=6*2 =12.

12 can be divided by 6 and 12.

Check with given numbers,

A, C and D are not available. So, C is an answer. If there is an E which is non of these. We have to choose it to be exact b/c we can use any number in a place of n. If so, 6, 12 and 18 are available.

Given expression is (6n^2 + 6n) = 6n(n+1).

Put n = 1.

Then,

= 6n(n+1).
= 6(1)*(1+1)
= 6*2 = 12.

Now 12 can be divided by 6 and 12.