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.

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

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.

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

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.

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

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

Priyanka said:   7 years ago
6n^2+6n= 6n(n+1).

Now the number above will always be divisible by 6
And it will be divisible by 12 only when n(n+1) part is even
Eg.6*2=12 ,6*6=36,6*5=30 when we used odd number to multiply with 6 its resultant was not divisible by 12 whereas with even number multiplies by 6 is always divisible by 12.

In case of n*(n+1) always one of the numbers will be odd and another one will be even, Multiplication of odd and even number always results into even number hence it will always be divisible by 12.

Sid said:   7 years ago
@Niranjan.

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

Narra Rakesh said:   6 years ago
If N is even number given expression is divisible by 6 and 12.
But, here N is a natural number.
The given expression is divisible by 6 only.


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