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# M13-30

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:50
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Difficulty:

25% (medium)

Question Stats:

69% (00:46) correct 31% (00:56) wrong based on 100 sessions

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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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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.

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19 Aug 2016, 05:06
why not 0,2,4???? then it must be C. Thans!!
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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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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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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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Joined: 13 Nov 2014
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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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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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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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15 Apr 2017, 02:37
wwambugu14 wrote:
Kindly show how you have moved from (2k−2)2k(2k+2) to 8(k−1)k(k+1)

Factor out 2 both from (2k - 2) and (2k + 2):

(2k − 2)2k(2k + 2) = 2*(k - 1)*2k*2*(k + 1) = 8*(k−1)k(k+1).

Hope it's clear.
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Re: M13-30   [#permalink] 15 Apr 2017, 02:37
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# M13-30

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