Aptitude - Numbers - Discussion

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.

If 46N is divisible by 18 then what is the value of N?

The numbers should be 6, 12, 18, 24.

Substitute any number for n.

It will become even number.

Hey I don't understand this problem please help!

Some books write it wrongly 6n2+6n instead of 6n^2 + 6n.

@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.

What is mean product of even, odd is always even?

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.

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.