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If an integer n is divisible by both 6 and 8, then it must

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Director
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If an integer n is divisible by both 6 and 8, then it must [#permalink] New post 04 Jul 2007, 19:50
If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

Though I got the correct ans by brute force. Just wondering which math rule could be applied here.
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Re: Math rule ? [#permalink] New post 04 Jul 2007, 23:12
Amit05 wrote:
If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

Though I got the correct ans by brute force. Just wondering which math rule could be applied here.


The way I figure it is taking the product of the prime factors which are not common among the two.

6= 3*2
8= 2*2*2

prod. of prime factors not common = 2*2*3=12

Answer B
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 [#permalink] New post 04 Jul 2007, 23:13
Find the prime factors of both, then multiply.

Ex)

6=(2)(3)
8=(2)(2)(2)

LCM = 24

24 is only divisible by 12. Answer is B.
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Re: Math rule ? [#permalink] New post 05 Jul 2007, 05:07
GK_Gmat wrote:
Amit05 wrote:
If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

Though I got the correct ans by brute force. Just wondering which math rule could be applied here.


The way I figure it is taking the product of the prime factors which are not common among the two.

6= 3*2
8= 2*2*2

prod. of prime factors not common = 2*2*3=12

Answer B


I like this explanation. If it's divisible by 6, it must have every prime 6 has, and the same thing is true for 8. So there must be 2,2,3. Even if you don't find the lowest common multiple, as in above, you should still just look for the answers with just some 2's and one 3. 12 is the only one.

Note that 18 doesn't work because there's an extra 3 in it.

Last edited by ian7777 on 05 Jul 2007, 07:12, edited 1 time in total.
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 [#permalink] New post 05 Jul 2007, 06:17
I would go with the LCM approach since i believe it is more feasible when dealing with unusual numbers.
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 [#permalink] New post 05 Jul 2007, 07:11
METHOD 1

6
0, 6, 12, 18, 24

8
0, 8, 16, 24

LCM = 24 which is divisible by 12.
hence, B.

METHOD 2

I also like the factoring approach for smaller numbers.

6= 2.3
8= 2.2.2

therefore, 2.2.3= 12, hence, B.
  [#permalink] 05 Jul 2007, 07:11
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If an integer n is divisible by both 6 and 8, then it must

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