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Is a*b*c divisible by 24? (1) a,b, and c are consecutive

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Is a*b*c divisible by 24? (1) a,b, and c are consecutive [#permalink]

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New post 28 May 2012, 05:06
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Is a*b*c divisible by 24?

(1) a, b, and c are consecutive even integers
(2) a*b is divisible by 12

This problem is from GMATClub Quant tests.
I am confused in this problem. Since 0 is also an EVEN number so statement 1 cannot be true. Am i right?
Please help!
[Reveal] Spoiler: OA

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Last edited by Bunuel on 29 May 2012, 00:39, edited 1 time in total.
Edited the question and added the OA
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Re: GMATClub number problem confusion [#permalink]

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New post 28 May 2012, 07:45
Is the answer A?

Even if one of the even integers a, b, c is 0, the product a*b*c = 0 is divisible by 24.

0/24 = 0 => integer. So a * b * c is divisible by 24.
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Joined: 12 Mar 2012
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Location: India
Concentration: Technology, General Management
GMAT Date: 07-23-2012
WE: Programming (Telecommunications)
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Kudos [?]: 44 [0], given: 4

Re: GMATClub number problem confusion [#permalink]

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New post 28 May 2012, 08:42
Oops! here is the catch which you pointed out. 0 is divisible by 24. Now I got it.
Yes the answer is A. Thanks a lot for the explanation.
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Re: Is a*b*c divisible by 24? (1) a,b, and c are consecutive [#permalink]

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New post 29 May 2012, 00:40
Expert's post
dexerash wrote:
Is a*b*c divisible by 24?

(1) a, b, and c are consecutive even integers
(2) a*b is divisible by 12

This problem is from GMATClub Quant tests.
I am confused in this problem. Since 0 is also an EVEN number so statement 1 cannot be true. Am i right?
Please help!


Note that an integer \(a\) is a multiple of an integer \(b\) (integer \(a\) is a divisible by an integer \(b\)) means that \(\frac{a}{b}=integer\): so, as 0 divided by any integer (except zero itself) yields an integer then zero is a multiple of every integer (except zero itself).

Also on GMAT when we are told that \(a\) is divisible by \(b\) (or which is the same: "\(a\) is multiple of \(b\)", or "\(b\) is a factor of \(a\)"), we can say that:
1. \(a\) is an integer;
2. \(b\) is an integer;
3. \(\frac{a}{b}=integer\).

BACK TO THE ORIGINAL QUESTION:
Is a*b*c divisible by 24?

(1) a, b, and c are consecutive even integers --> \(a=2k-2\), \(b=2k\) and \(c=2k+2\) for some integer \(k\) --> \(abc=(2k-2)2k(2k+2)=8(k-1)k(k+1)\), 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. 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.

(2) a*b is divisible by 12, clearly insufficient as no info about c (if ab=12 and c=1 answer will be NO but if ab=24 and c=any integer then the answer will be YES).

Answer: A.

Hope it's clear.
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Re: Is a*b*c divisible by 24? (1) a,b, and c are consecutive   [#permalink] 29 May 2012, 00:40
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Is a*b*c divisible by 24? (1) a,b, and c are consecutive

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