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Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
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.

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
14 Feb 2015, 14:47

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Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
14 Feb 2015, 20:00

Expert's post

Hi All,

In these sorts of situations, prime-factorization is a great technical approach to get to the correct answer. There is another method that is actually pretty easy though - we can TEST VALUES. Since the prompt asks for what N MUST be divisible by, we just need to start with the SMALLEST positive integer for N that is DIVISIBLE by BOTH 6 and 8.

Many Test Takers would say that 48 is the smallest integer, but it's NOT. The smallest integer is actually 24. Here's proof that's fairly easy to put together....

Multiples of 6: 6, 12, 18, 24 Multiples of 8: 8, 16, 24

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