divisibility

One of the easiest(if slightly cumbersome) ways to solve this is to substitute a coupla values and see what you get.

Try values like 10, 14, 22

It can get done in the 1.5mins that you allocate for each problem

Ans is E rt

nik08 wrote:If an even number n is not divisible by 3 or 4, then what must (n+6)(n+8 )(n+10) be divisible by ?
I. 24
II. 32
III. 96

A. None
B. I only
C. II only
D. I and II only
E. I , II and III

Can someone please help ?

thanks !

n is an even number not divisible by 3 or 4
(n+6)(n+8 )(n+10) represent the product of 3 consecutive even integers.

any 3 consecutive even integers will be divisible two 4's and one 3 at least
i.e. 4*4*3 = 16*3

24 = 8*3
32 = 8*4
96 = 8*4*3

Therefore the answer is E.
Hope this helps.

I don't know whether my approach is correct

So...
If an even number n is not divisible by 3 or 4, then what must (n+6)(n+8 )(n+10) be divisible by ?
I. 24
II. 32
III. 96

If an even number n is not divisible by 3 or 4
This means that we should take all even numbers that are not multiplies of 4 (2*2) and 3
let n = 10
so we have (n+6)(n+8 )(n+10) =16*18*20 = 2^4*2*3^2*2^2*5 = 2^7*3^2*5

factorising 24 = 2^3*3. 2^7*3^2*5 is divisible by 2^3*3
32 = 2^5. 2^7*3^2*5 is divisible by 2^5
96 = 2^5*3. 2^7*3^2*5 is divisible by 2^5*3

So answer E is correct

I cheked the answer using the same approach by using 14, 22 and 26 as n. This was the long process, but I think that if the smallest numbers can give the answer means that all bigger numbers will give the same answer. Because further increasing will give only additional prime factors and we will always have necessary factors - 2^5*3^2
remember that n is even, so we will always have three 2's

Thanks a lot guys ... i think the trial n err method wud take up a lot of time ..