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M13-30

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M13-30 [#permalink]

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New post 16 Sep 2014, 00:50
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If \(a\), \(b\) and \(c\) are positive integers is \(a*b*c\) divisible by 24?


(1) \(a\), \(b\), and \(c\) are consecutive even integers

(2) \(a*b\) is divisible by 12

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Re M13-30 [#permalink]

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New post 16 Sep 2014, 00:50
Official Solution:


(1) \(a\), \(b\), and \(c\) are consecutive even integers. Given: \(a=2k-2\), \(b=2k\) and \(c=2k+2\) for some integer \(k\), so \(abc=\) \((2k-2)2k(2k+2)=8(k-1)k(k+1)\), now \((k-1)\), \(k\), \((k+1)\) are 3 consecutive integers, which means that one of them must be a multiple of 3, thus \(abc\) is divisible by both 8 and 3, so by 24 too. Sufficient.

Or even without the formulas: the product of 3 consecutive even integers will have \(2*2*2=8\) as a factor, plus out of 3 consecutive even integers one must be a multiple of 3, thus \(abc\) is divisible by both 8 and 3, so by 24 too.

(2) \(a*b\) is divisible by 12, clearly insufficient as we don't have any information about \(c\) (if \(ab=12\) and \(c=1\), the answer will be NO but if \(ab=24\) and \(c=\text{any integer}\), then the answer will be YES). Not sufficient.


Answer: A
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Re: M13-30 [#permalink]

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New post 19 Aug 2016, 05:06
why not 0,2,4???? then it must be C. Thans!!
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Re: M13-30 [#permalink]

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New post 19 Aug 2016, 05:21
empo01 wrote:
why not 0,2,4???? then it must be C. Thans!!


ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html
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Re: M13-30 [#permalink]

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New post 27 Oct 2016, 05:50
What if the integers -4, -2, 0, then statement (1) is insufficient
But in case A*B is divisible by 12 like 2*4*6 so statement (2) alone is sufficient
Right ?
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Re: M13-30 [#permalink]

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New post 27 Oct 2016, 05:55
Emaco wrote:
What if the integers -4, -2, 0, then statement (1) is insufficient
But in case A*B is divisible by 12 like 2*4*6 so statement (2) alone is sufficient
Right ?


Not right. Please check the post just above yours: 0 is divisible by EVERY integer except 0 itself. As for (2): we don't know whether a, b, and c are consecutive for (2).
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Re: M13-30 [#permalink]

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New post 27 Oct 2016, 07:15
Correct solution is A

0 cannot be one of the integers in A because 0 is neither positive nor negative, which the stem states that the integers are all positive.

That means that a*b*c will be in the form of
2x*2x+2*2x+4. this is divisible by 24 whenever x >0

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Re: M13-30 [#permalink]

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New post 27 Oct 2016, 08:30
Emaco wrote:
What if the integers -4, -2, 0, then statement (1) is insufficient
But in case A*B is divisible by 12 like 2*4*6 so statement (2) alone is sufficient
Right ?


Question says:

If \(a\), \(b\) and \(c\) are positive integers is \(a*b*c\) divisible by 24?

That eliminates zero and negative numbers.

Anyway, 0 is also divisible by 12.
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Re: M13-30 [#permalink]

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New post 15 Apr 2017, 02:30
Kindly show how you have moved from (2k−2)2k(2k+2) to 8(k−1)k(k+1)
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Re: M13-30 [#permalink]

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New post 15 Apr 2017, 02:37
Re: M13-30   [#permalink] 15 Apr 2017, 02:37
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