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

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05 Jul 2007, 06:07

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, 08:12, edited 1 time in total.

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

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14 Feb 2015, 15:47

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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

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

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24 Feb 2016, 17:55

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

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27 May 2016, 05:45

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

A very straightforward method involving the LCM exists to solve the problem.

First of all, "n" is divisible by both 6 and 8; This implies "n" is a Multiple of both 6 and 8.

Finding the LCM helps as all other Multiples of 6 and 8 may have extra prime factors which may lead to "n" be divisible by other numbers.

e.g. the LCM of 6 and 8 is 24, but if we find any multiple of 6 and 8 such as 48 or 72 then apart from 12, the number 48 is also divisible by 16 and 72 by 18.

Finally, check the divisiblity of the LCM by the options provided. Only one value should satisfy the divisibility condition.

Cheers!!!... Press "Kudos" if you liked the explanation.
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Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]

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01 Sep 2017, 03:53

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

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02 Sep 2017, 08:57

n is divisible by both 6 and 8 means .... n = 6*8

n = 6*8 = (2^4) * 3 implies that the answer must meet two conditions: (1) be divisible by the product of its prime which is 6=2*3. This by itself narrows down the answer to B=12 or E=18. (2) contains at most one 3 and at most four 2's. E is not correct because it factors into two 3's which is one too many 3's.

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

Let’s first find the LCM of 6 and 8. The prime factorization of 6 is 3 x 2 and the prime factorization of 8 is 2 x 2 x 2. Thus, the LCM of 6 and 8 is 3 x 2 x 2 x 2 = 24. Of all the answer choices, only 12 is a factor of 24, so n is divisible by 12.

Answer: B
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