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Is integer x divisible by 12?

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s0cck4z
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Is integer x divisible by 12? [#permalink] New post   28 Nov 2015, 09:08
This is a question about one of the statements, NOT about how to answer the question itself. The problem is taken from Veritas, Book 8--Data sufficiency.

A. x is the product of three consecutive positive integers

B. x is the product of three consecutive even integers

C. x is the product of three consecutive prime numbers

D. x^2 is divisible by 36

E. x^2 is divisible by 72

Regarding the bolded statement: Veritas says this statement is sufficient and proves it using numbers greater than 0, but couldn't you say 0*2*4?
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Re: Is integer x divisible by 12? [#permalink] New post   28 Nov 2015, 09:13
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s0cck4z wrote:
This is a question about one of the statements, NOT about how to answer the question itself. The problem is taken from Veritas, Book 8--Data sufficiency.

A. x is the product of three consecutive positive integers

B. x is the product of three consecutive even integers

C. x is the product of three consecutive prime numbers

D. x^2 is divisible by 36

E. x^2 is divisible by 72

Regarding the bolded statement: Veritas says this statement is sufficient and proves it using numbers greater than 0, but couldn't you say 0*2*4?


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: Is integer x divisible by 12? [#permalink] New post   28 Nov 2015, 11:38
What are the constraints on x?
0*2*4? = 0
As Bunuel mentioned, 0 is an integer. So if the requirement is that x has to be an integer, well the above still holds true.
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Re: Is integer x divisible by 12? [#permalink] New post   28 Nov 2015, 11:49
GMATPill wrote:
What are the constraints on x?
0*2*4? = 0
As Bunuel mentioned, 0 is an integer. So if the requirement is that x has to be an integer, well the above still holds true.


The only constraint on x was that it be an integer.
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Re: Is integer x divisible by 12? [#permalink] New post   28 Nov 2015, 14:00
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Hi s0cck4z,

Yes, you could have used 0, 2 and 4. ANY 3 consecutive integers (even if you use negatives or 0) will yield a product that is divisible by 12.

eg.
(0)(2)(4) = 0 which is divisible by 12.
(2)(4)(6) = 48 which is divisible by 12.

GMAT assassins aren't born, they're made,
Rich
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Is integer x divisible by 12? [#permalink] New post   28 Nov 2015, 15:17
EMPOWERgmatRichC wrote:
Hi s0cck4z,

Yes, you could have used 0, 2 and 4. ANY 3 consecutive integers (even if you use negatives or 0) will yield a product that is divisible by 12.

eg.
(0)(2)(4) = 0 which is divisible by 12.
(2)(4)(6) = 48 which is divisible by 12.

GMAT assassins aren't born, they're made,
Rich


Thanks, everyone! If you're in the US, hope you had a great Thanksgiving--and if you aren't, well, cheers to you just the same!
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Integers [#permalink] New post   30 Sep 2020, 16:33
This is a question about one of the statements, NOT about how to answer the question itself. The problem is taken from Veritas, Book 8--Data sufficiency.

Is the Integer X divisible by 12

A. x is the product of three consecutive positive integers

B. x is the product of three consecutive even integers

C. x is the product of three consecutive prime numbers

D. x^2 is divisible by 36

E. x^2 is divisible by 72

Can someone explain why D is insufficient and E is Sufficient as I am not satisfied with the answer explanation in the book.
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Re: Integers [#permalink] New post   30 Sep 2020, 19:17
For D: If you take x=6, then x^2 would be 36, in this case, x won't be divisible by 12.

For E: for x² to be divisible by 72(2³*3²), x would have to be atleast divisible by (2²*3). Hence it is sufficient.
Did that make sense?

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Re: Integers [#permalink] New post   30 Sep 2020, 21:16
Is the Integer X divisible by 12

D. x^2 is divisible by 36 : 36 = 2^2 * 3 ^2 * k => x = 6k => x can be 6, 12, 18 etc

So, not necessarily divisible by 12 ( 6, 18 etc)

E. x^2 is divisible by 72 : 72 = 2^3 * 3 ^2 * k => x = 2^2 ( since it is x^2, the power of 2 must be even in x^2 (so we take it as 4 rather than 3) * 3 or 12k => Hence E is the answer
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Re: Integers [#permalink]
30 Sep 2020, 21:16
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