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divisibility

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by nik08 » Sat Sep 06, 2008 6:19 am
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 !
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by Neo2000 » Sat Sep 06, 2008 6:47 am
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
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by kshankker » Sat Sep 06, 2008 7:45 am
Ans is E rt
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Re: divisibility

by parallel_chase » Sat Sep 06, 2008 10:01 am
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.
Last edited by parallel_chase on Sat Sep 06, 2008 12:55 pm, edited 2 times in total.
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by 4meonly » Sat Sep 06, 2008 10:13 am
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
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by nik08 » Sun Sep 07, 2008 2:10 am
Thanks a lot guys ... i think the trial n err method wud take up a lot of time ..
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by Ian Stewart » Sun Sep 07, 2008 8:20 am
If an even number n is not divisible by 3 or 4, then what must (n+6)(n+8 )(n+10) be divisible by ?

We know n is even, but not divisible by 4. If you look at consecutive even numbers:

2, 4, 6, 8, 10, 12, 14, 16, ...

every second even number is a multiple of 4. So if n is not a multiple of 4, n+2 must be, as must be n+6 and n+10. That is, n+6 and n+10 are multiples of 4 (and one of them must actually be a multiple of 8, but we don't need that here), and n+8 is a multiple of 2. Thus

(n+6)(n+8)(n+10) must be a multiple of 4*2*4 = 32.

In addition, n+6, n+8, and n+10 are three consecutive even numbers. Exactly one of them must be a multiple of 3, since every third even number is a multiple of 3. So

(n+6)(n+8)(n+10) must be a multiple of 4*2*4*3 = 96.

Every multiple of 96 is also a multiple of 32 and 24, so E.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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