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n is not divisible by 3 or 4

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by neerajkumar1_1 » Wed Jun 23, 2010 8:18 pm
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

OA:E

I have been able to get the answer and I have done it just over 2 mins. I solved the question by realizing 2, 10 and 14 will not be divisble by 2 & 3 and then factorizing (n + 6)(n + 8)(n + 10) to find the common factors in 24,32 and 96.

Now when i was looking at the explanation i found it a better way to think:
So here is the explanation:

If n is an even number not divisible by 4, then n + 8 will also be an even number not divisible by 4.
However, n + 6 and n + 10 will both be even numbers that are divisible by 4.
Furthermore, n + 6 and n + 9 wouldn't be divisible by 3 either, but either n + 8 or n +10 must be divisible be 3
because at least one of three consecutive integers must be divisible by 3.
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


I find the method pretty good, but I am unable to understand the last step in which it mentions:
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


Here I get 4 x 4 x 3..... but i dont understand how according to their logic they also add 2 in the must be divisors.

Please let me know your views...

Thanks...
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by ayushiiitm » Wed Jun 23, 2010 8:36 pm
and how did we get 4*4*3


confusion method...need clarity
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by Rich@VeritasPrep » Wed Jun 23, 2010 8:47 pm
Hey Neeraj,

While it is well and good to search for thorough algebraic explanations when you have the time to do so, it is not very helpful when you have only 2 minutes average to answer a question.

This particular question is one solved much more easily with testing numbers. You are searching for what the product MUST BE divisible by. The best strategy, therefore, is to choose the lowest possible value of n, so you have the fewest possible number of divisors of (n+6)(n+8)(n+10).

By generating the fewest possible number of divisors, you are being more restrictive, which means you will find out how many of the three given options (24, 23, 96) MUST ALWAYS be factors of the product.

In this case, the smallest possible value of n is 2. In that case, the product is (2+6)(2+8)(2+10) = 960.

960 is divisible by 24, 32, and 96; therefore, the answer is E.

This might not be the most complete approach to solving all of the algebraic properties to this question, but your goal is just to find the correct answer. That is all you should be concerned about.

Hope this helps!
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by amising6 » Wed Jun 23, 2010 10:25 pm
neerajkumar1_1 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

OA:E

n is an even number so n+6 will be an evn number n+8 will also be even number n+10 will also be even number
if you look at it carefully its 3 continious even number .and n should not be divisible by 3 or 4 so let us take one smallest value of n i.e 0 or for sake 2 i.e
if n=2

8*10*12=2*2*2*2*5*2*2*3 so it will contain 24 ,32 ,96 all of thm so i guess answer will be E
you cant take 0 as 0 isdivsible by 4 i.e 0/4=0
een able to get the answer and I have done it just over 2 mins. I solved the question by realizing 2, 10 and 14 will not be divisble by 2 & 3 and then factorizing (n + 6)(n + 8)(n + 10) to find the common factors in 24,32 and 96.

Now when i was looking at the explanation i found it a better way to think:
So here is the explanation:

If n is an even number not divisible by 4, then n + 8 will also be an even number not divisible by 4.
However, n + 6 and n + 10 will both be even numbers that are divisible by 4.
Furthermore, n + 6 and n + 9 wouldn't be divisible by 3 either, but either n + 8 or n +10 must be divisible be 3
because at least one of three consecutive integers must be divisible by 3.
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


I find the method pretty good, but I am unable to understand the last step in which it mentions:
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


Here I get 4 x 4 x 3..... but i dont understand how according to their logic they also add 2 in the must be divisors.

Please let me know your views...

Thanks...

see the bold part
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by kvcpk » Thu Jun 24, 2010 12:50 am
I was finally able to find the algebraic solution for this problem.. Though not important from GMAT perspective.. Its good to know the approach..
Here it comes..

Let the even number be 2k.

We need to find what (n + 6)(n + 8)(n + 10) is divisible by:
Substituting n=2k, we get,
(2k + 6)(2k + 8)(2k + 10)
= 8(k+ 3)(k+ 4)(k+ 5)

Given that 2k is not divisible by 3 or 4.
So 2k should be of the form:
3a +1 or
3a + 2 or
4b + 1 or
4b + 2 or
4b + 3

From the above forms let us substitute 3a+1 for value of k

we get, 8(3a+4)(3a+5)(3a+ 6)
=24(3a+4)(3a+5)(a+ 2)
So the result is divisible by 24.

We get the same if we substitute 3a+2 also...

Substituting 4b+1 in place of k, we get,
8(4b+4)(4b+5)(4b+ 6)
=32(b+1)(4b+5)(4b+ 6)
So the result is divible by 32.

If a value is divisible by both 24 and 32, then the value is divisble by 96 also. Because 96 is LCM of 24,32

Hence answer is E.

Hope this helps!!
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by sanju09 » Thu Jun 24, 2010 1:03 am
neerajkumar1_1 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

OA:E

I have been able to get the answer and I have done it just over 2 mins. I solved the question by realizing 2, 10 and 14 will not be divisble by 2 & 3 and then factorizing (n + 6)(n + 8)(n + 10) to find the common factors in 24,32 and 96.

Now when i was looking at the explanation i found it a better way to think:
So here is the explanation:

If n is an even number not divisible by 4, then n + 8 will also be an even number not divisible by 4.
However, n + 6 and n + 10 will both be even numbers that are divisible by 4.
Furthermore, n + 6 and n + 9 wouldn't be divisible by 3 either, but either n + 8 or n +10 must be divisible be 3
because at least one of three consecutive integers must be divisible by 3.
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


I find the method pretty good, but I am unable to understand the last step in which it mentions:
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


Here I get 4 x 4 x 3..... but i dont understand how according to their logic they also add 2 in the must be divisors.

Please let me know your views...

Thanks...

The smallest n fulfilling the need is 2, which gives the smallest (n + 6) (n + 8) (n + 10) as equal to 8 × 10 × 12; all 24, 32, and 96 are a factor of this smallest possibility.

[spoiler]E[/spoiler]
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by jube » Thu Jun 24, 2010 3:44 am
raz1024 wrote:Hey Neeraj,

While it is well and good to search for thorough algebraic explanations when you have the time to do so, it is not very helpful when you have only 2 minutes average to answer a question.

This particular question is one solved much more easily with testing numbers. You are searching for what the product MUST BE divisible by. The best strategy, therefore, is to choose the lowest possible value of n, so you have the fewest possible number of divisors of (n+6)(n+8)(n+10).

By generating the fewest possible number of divisors, you are being more restrictive, which means you will find out how many of the three given options (24, 23, 96) MUST ALWAYS be factors of the product.

In this case, the smallest possible value of n is 2. In that case, the product is (2+6)(2+8)(2+10) = 960.

960 is divisible by 24, 32, and 96; therefore, the answer is E.

This might not be the most complete approach to solving all of the algebraic properties to this question, but your goal is just to find the correct answer. That is all you should be concerned about.

Hope this helps!
I have a doubt about this approach. While (2+6)(2+8)(2+10) is divisble by all the nos., how do we know for sure whether another no. say 5 or 10 or 62 or any other no. which satisfies the condition will also necessarily be divisble by all the nos.

thanks!
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by kvcpk » Thu Jun 24, 2010 3:49 am
jube wrote:
I have a doubt about this approach. While (2+6)(2+8)(2+10) is divisble by all the nos., how do we know for sure whether another no. say 5 or 10 or 62 or any other no. which satisfies the condition will also necessarily be divisble by all the nos.

thanks!
Hi Jube,

You can have a look at my post above for detailed explanation..
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by jube » Thu Jun 24, 2010 3:50 am
kvcpk wrote:I was finally able to find the algebraic solution for this problem.. Though not important from GMAT perspective.. Its good to know the approach..
Here it comes..

Let the even number be 2k.

We need to find what (n + 6)(n + 8)(n + 10) is divisible by:
Substituting n=2k, we get,
(2k + 6)(2k + 8)(2k + 10)
= 8(k+ 3)(k+ 4)(k+ 5)

Given that 2k is not divisible by 3 or 4.
So 2k should be of the form:
3a +1 or
3a + 2 or
4b + 1 or
4b + 2 or
4b + 3
quick question about the above - most likely I'm missing something so will be awesome if you could clarify.

If we take 2k to be 3n + 1 then won't 4 also figure in there? similarly, if we take it to be of the form 3n + 2 then won't 8 also figure in the list?

thanks!
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by kvcpk » Thu Jun 24, 2010 4:09 am
jube wrote: quick question about the above - most likely I'm missing something so will be awesome if you could clarify.

If we take 2k to be 3n + 1 then won't 4 also figure in there? similarly, if we take it to be of the form 3n + 2 then won't 8 also figure in the list?

thanks!
Good question.. the point here is we are just trying to mimic the number sequence structure.. Any number as per the question should be of any of those 5 forms.. So we just need to see if it is working for all the 5 forms of numbers..

Infact the 5 series should be like this:

(3a+1 or 3a +2) and (4b+1 or 4b+2 or 4b +3)
So this eliminates the possibility of it being a multiple of 4.

Hope this helps!!.. Let me know if you have any queries..
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by Stuart@KaplanGMAT » Thu Jun 24, 2010 1:17 pm
neerajkumar1_1 wrote:
If n is an even number not divisible by 4, then n + 8 will also be an even number not divisible by 4.
However, n + 6 and n + 10 will both be even numbers that are divisible by 4.
Furthermore, n + 6 and n + 9 wouldn't be divisible by 3 either, but either n + 8 or n +10 must be divisible be 3
because at least one of three consecutive integers must be divisible by 3.
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


I find the method pretty good, but I am unable to understand the last step in which it mentions:
All together, we find that the product (n + 6)(n + 8)(n + 10) will be divisible by 2 × 4 × 4 × 3.


Here I get 4 x 4 x 3..... but i dont understand how according to their logic they also add 2 in the must be divisors.

Please let me know your views...

Thanks...
To answer the original question (which I don't think has been addressed):

both (n+6) and (n+10) are divisible by 4 (those are our 2 "4"s); since n is even, even though (n+8) isn't a multiple of 4, it is a multiple of 2 (that's our "2").
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by pradeepkaushal9518 » Fri Jun 25, 2010 12:54 am
how i have solved even numbers not divisible by 3 or 4 are 2, 10,14....so on

taking 2 (n+6)(n+8)(n+10) = 8*10*12 which is divisible by 24,32 and 96 also hence i,ii and iii so answer is E

i dont think this question really need that much attentions and explanations
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by kevincanspain » Fri Jun 25, 2010 1:16 am
raz1024 wrote:Hey Neeraj,

While it is well and good to search for thorough algebraic explanations when you have the time to do so, it is not very helpful when you have only 2 minutes average to answer a question.

This particular question is one solved much more easily with testing numbers. You are searching for what the product MUST BE divisible by. The best strategy, therefore, is to choose the lowest possible value of n, so you have the fewest possible number of divisors of (n+6)(n+8)(n+10).

By generating the fewest possible number of divisors, you are being more restrictive, which means you will find out how many of the three given options (24, 23, 96) MUST ALWAYS be factors of the product.

In this case, the smallest possible value of n is 2. In that case, the product is (2+6)(2+8)(2+10) = 960.

960 is divisible by 24, 32, and 96; therefore, the answer is E.

This might not be the most complete approach to solving all of the algebraic properties to this question, but your goal is just to find the correct answer. That is all you should be concerned about.

Hope this helps!
Careful! Choosing the smallest possible value of n does not always generate the smallest number of divisors. If you want to pick numbers, it is good to look at consecutive examples. Learning the basics about divisibility is well worth the effort, and with a bit of practice, this question can be done in 30 seconds
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