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.

NIRANJAN said:   7 years ago
If 46N is divisible by 18 then what is the value of N?

Prudvi said:   8 years ago
The numbers should be 6, 12, 18, 24.

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

It will become even number.

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

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

Vipul said:   1 decade ago
@Emi.

It mean that --** 6n (n+1) **--- where 6 is product & n may be even or odd.

That's why if product is even is always give even result whether -- n -- may have even or odd.

Emi said:   1 decade ago
What is mean product of even, odd is always even?

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.

Khin Thiri Htet said:   1 decade ago
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.

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

540/12 = 45.


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