Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 28)
28.
If n is a natural number, then (6n2 + 6n) is always divisible by:
6 only
6 and 12 both
12 only
by 18 only
Answer: Option
Explanation:

(6n2 + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.

Discussion:
27 comments Page 2 of 3.

Priya said:   1 decade ago
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?

Deepak maurya said:   1 decade ago
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.

Shruti said:   3 years ago
If we assume that n = 1.
Then,
6n^2+6n,
6*1*1+6*1 = 12,
So from the given options, 6 and 12 is always divisible.
(3)

Sid said:   7 years ago
@Niranjan.

It would be 8,
because 18x25=450,
And then so, 18x26=468,
So, N would be 8.

Onkar said:   5 years ago
When N is odd. Then it's not divisible by 12.

So, it's divisible by only 6.

K.pavan said:   1 decade ago
ELIMINATION METHOD...

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

Hussain said:   1 decade ago
Who told you that 540 is not divisible by 12?

540/12 = 45.

VISITHRA .S said:   8 years ago
I do not understand this problem. Please explain in detail.

Jayaprasath said:   8 years ago
Substitute any number for n.

It will become even number.

Veer said:   9 years ago
Some books write it wrongly 6n2+6n instead of 6n^2 + 6n.


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