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M13 Q30

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M13 Q30 [#permalink] New post 03 Sep 2008, 18:02
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The question is
Is a*b*c divisible by 24?

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

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

Answer given is:
One of any three consecutive even integers is divisible by 3. Because this integer is even, it is also divisible by 6. When multiplied by two more even integers, it renders a product that is divisible by 24. S_1 is sufficient. S_2 is not sufficient. We need to know something about c.

The correct answer is A.

My question
What about 0? i.e. 0, 2, 4, shouldn't the answer be E?
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Re: M13 Q30 [#permalink] New post 24 Dec 2010, 06:51
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Tadashi wrote:
The question is
Is a*b*c divisible by 24?

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

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

Answer given is:
One of any three consecutive even integers is divisible by 3. Because this integer is even, it is also divisible by 6. When multiplied by two more even integers, it renders a product that is divisible by 24. S_1 is sufficient. S_2 is not sufficient. We need to know something about c.

The correct answer is A.

My question
What about 0? i.e. 0, 2, 4, shouldn't the answer be E?


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 yes, 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: th 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: M13 Q30 [#permalink] New post 17 Dec 2009, 08:11
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Another way of phrasing the question would be: is abc/24 an integer?

Recall that an integer is a whole number in the infinite set {. . . -3, -2, -1, 0, 1, 2, 3, . . .}.

Statement 2:
If abc is divisible by 12, then it could also be divisible by 24 if the the number equalled 24, 48, 240, etc. But if abc is 12, then the number is not divisible by 12. In short, we can't determine the answer. NOT SUFFICIENT.

Statement 1:
Let's start with the smallest absolute values allowed first: 0, 2, and 4. When multiplied, this equals zero - an integer. When zero is divided by any non-zero integer, the result is zero - again, an integer. So it's divisible by 24. If we use, negative numbers -2, -4, -6, we arrive at -48/24 = -2, an integer. If we break down 24 into primes, we have one 3 and three 2's. Pick any consecutive set of even integers. These prime numbers will be in there. In other words the number is divisible by 24. SUFFICIENT.


And the GMAT assumes that you know about the divisibility rules of 0, just as it assumes that you know about the rules for 3, 2, 10, 5, etc.
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Re: M13 Q30 [#permalink] New post 03 Sep 2008, 23:29
Hi. You forget that 0 is divisible by every number except for itself. See this link for more info:

http://gmatclub.com/wiki/Math_definitions

Hope this helps.
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Re: M13 Q30 [#permalink] New post 04 Sep 2008, 16:09
Thanks, I did not know that 0 is divisible by every number excluding 0. It would have helped to have that in the explanation.
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m13 q30 [#permalink] New post 23 Feb 2009, 05:25
Hi all,

I've come across a question that I'm a bit confused by and would appreciate your help with.

Question:
Is a*b*c divisible by 24 ?

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

GMATClub Answer:
One of any three consecutive even integers is divisible by 3. Because this integer is even, it is also divisible by 6. When multiplied by two more even integers, it renders a product that is divisible by 24. S_1 is sufficient.

S_2 is not sufficient. We need to know something about c.

The correct answer is A.

-----

The GMATClub answer seems to be assuming that the consecutive integers are positive. Is this not an incorrect assumption? The question doesn't state the even integers are positive.

As an example: the values for a, b and c could be 2, 4 and 0, whose product is not divisible by 24. Or the values could be 2, 4 and 6, for which the product is divisible by 24. So S1 seems insufficient.


Could someone let me know if I've misunderstood this? Many thanks in advance,

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Re: M13 Q 30 - Data Sufficiency & Number Properties [#permalink] New post 23 Feb 2009, 07:33
I think 0 is divisible by any number... not sure though and -ve numbers I dont think make a difference here as only the sign will change but the number will remain divisible by 24.
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Re: M13 Q30 [#permalink] New post 24 Dec 2010, 06:35
A
24 prime factors set = 2*2*2*3
the product of three consecutive even integers do contain 2*2*2*3 hence answers is 1 is sufficient

if a*b is divisible 12? nothing mentioned about c. So not sufficient.
Answers is A.
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Re: M13 Q30 [#permalink] New post 27 Dec 2010, 04:56
+1 for A.

(a) 24 = 3 * 8 = 3 * 2 * 4 = 2 * 3 * 4

If you take any 3 consecutive 3 integers, they have 2, 3 and 4 as multiples!

(b) No information about c.
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Re: M13 Q30 [#permalink] New post 28 Dec 2010, 14:40
Tadashi wrote:
The question is
Is a*b*c divisible by 24?

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

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

Answer given is:
One of any three consecutive even integers is divisible by 3. Because this integer is even, it is also divisible by 6. When multiplied by two more even integers, it renders a product that is divisible by 24. S_1 is sufficient. S_2 is not sufficient. We need to know something about c.

The correct answer is A.

My question
What about 0? i.e. 0, 2, 4, shouldn't the answer be E?


This is how I went about solving the problem...
Statement 1)
I factored 24 and got 2, 2, 2, 3
I used a MGMAT strategy that an even x even = divisible by 4
even x even x even = divisible by 8
Since there are three consecutive even numbers I knew it was even x even x even (at least divisible by 2,2,2) and since there are three terms it has to be divisible by 3

Statement 2) is insufficient because there is no mention of C. If a*b = 12 and c is 1 then the answer yields NO. However, if we make a*b = 24 and c = 1 we get an answer of YES. Thus, insufficient.
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Re: M13 Q30 [#permalink] New post 26 Oct 2011, 03:52
The correct answer has to be C.

As per OG 12 - 0 is an even number.

Last edited by madzy on 26 Oct 2011, 04:17, edited 1 time in total.
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Re: M13 Q30 [#permalink] New post 26 Oct 2011, 04:04
Please see this post above:
m13-q30-69735.html#p841095

The answer is A.

madzy wrote:
The correct answer has to be E.

As per OG 12 - 0 is an even number.

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Re: M13 Q30 [#permalink] New post 30 Dec 2012, 21:07
Clearly, Statement 1 is Sufficient.Any three consecutive evern numbers will have 3 multiple of 2 that comes to 2*2*2 = 8.Also, any three consecutive evern numbers will have atleast 1 multiple of 3, so that leads to 8 (from above)*3=24.Hence, any three consecutive even no is divisible by 24.
If we take 0 as one of the even nos among the three, then also 0 is divisible by any no >0.Hence Statement 1 Is sufficient.

Statement 2 :- Insufficient, Imagine 3,4 and 1 as one of the sets that is divisible by12 but not by 24.Hence, Insufficient.
I would rate this question as 600 level and nothing beyond for sure.

Thanks,
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Re: M13 Q30 [#permalink] New post 02 Jan 2013, 09:56
BM wrote:
Another way of phrasing the question would be: is abc/24 an integer?

Recall that an integer is a whole number in the infinite set {. . . -3, -2, -1, 0, 1, 2, 3, . . .}.

Statement 2:
If abc is divisible by 12, then it could also be divisible by 24 if the the number equalled 24, 48, 240, etc. But if abc is 12, then the number is not divisible by 12. In short, we can't determine the answer. NOT SUFFICIENT.

Statement 1:
Let's start with the smallest absolute values allowed first: 0, 2, and 4. When multiplied, this equals zero - an integer. When zero is divided by any non-zero integer, the result is zero - again, an integer. So it's divisible by 24. If we use, negative numbers -2, -4, -6, we arrive at -48/24 = -2, an integer. If we break down 24 into primes, we have one 3 and three 2's. Pick any consecutive set of even integers. These prime numbers will be in there. In other words the number is divisible by 24. SUFFICIENT.


And the GMAT assumes that you know about the divisibility rules of 0, just as it assumes that you know about the rules for 3, 2, 10, 5, etc.



Regarding Statement 2:
since a*b is divisible by 12...e can write a*b as 12*X(considering X as integer)------(1).
since c is an even number c can be written as 2*Y(considering Y as integer)----------(2).

From (1) and (2) e can write a*b*c = 12*X*2*y=24XY .......so a*b*c is equal to 24XY hich is obviously divisible by 12

so i think statement 2 alone is also SUFFICENT.....Please correct me if iam wrong
Re: M13 Q30   [#permalink] 02 Jan 2013, 09:56
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