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S is a set of positive integers such that if integer X is a

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S is a set of positive integers such that if integer X is a [#permalink] New post 15 Aug 2007, 19:13
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S is a set of positive integers such that if integer X is a member of S, then both X^2 and X^3 are also in S. If the only member of S that is neither the square nor the cube of another member of S is called the Source Integer, is 8 in S?

1) 4 is in S and is not the Source Integer.
2) 64 is in S and is not the Source Integer.

---

This is a question from Kaplan. Pls explain.

Thanks.
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Re: A quick DS question [#permalink] New post 15 Aug 2007, 19:33
amitavha wrote:
S is a set of positive integers such that if integer X is a member of S, then both X^2 and X^3 are also in S. If the only member of S that is neither the square nor the cube of another member of S is called the Source Integer, is 8 in S?

1) 4 is in S and is not the Source Integer.
2) 64 is in S and is not the Source Integer.

---

This is a question from Kaplan. Pls explain.

Thanks.


Well I had time understanding the question but I will take the first stab.

1) Stmnt 1 - 4 is in S but not source so source has to be 2 and 8 will be in set . SUFF

2) Stmnt 2 - 64 is in set means there could be two possible combination
( 4 , 4^2 , 4^3 } or { 8 , 8^2 , 8^3 } . So seocnd set does has 8 but first set doesnt . So not SUFF because we cant determine for sure whther set has 8.

So I guess anwer is A
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Re: A quick DS question [#permalink] New post 15 Aug 2007, 21:15
amitavha wrote:
S is a set of positive integers such that if integer X is a member of S, then both X^2 and X^3 are also in S. If the only member of S that is neither the square nor the cube of another member of S is called the Source Integer, is 8 in S?

1) 4 is in S and is not the Source Integer.
2) 64 is in S and is not the Source Integer.

This is a question from Kaplan. Pls explain.

Thanks.



1: if 4 is in S and is not a source integer, 2 is there and subsequently 8 is there. so suff.........

2. if 64 is in set S, and 64 is not a source integer, 64 could have because of 4^3 or of 8^2. since 4 is there, 8 is also there. so again suff.....

should be D.
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Re: A quick DS question [#permalink] New post 15 Aug 2007, 21:28
Fistail wrote:
amitavha wrote:
S is a set of positive integers such that if integer X is a member of S, then both X^2 and X^3 are also in S. If the only member of S that is neither the square nor the cube of another member of S is called the Source Integer, is 8 in S?

1) 4 is in S and is not the Source Integer.
2) 64 is in S and is not the Source Integer.

This is a question from Kaplan. Pls explain.

Thanks.



1: if 4 is in S and is not a source integer, 2 is there and subsequently 8 is there. so suff.........

2. if 64 is in set S, and 64 is not a source integer, 64 could have because of 4^3 or of 8^2. since 4 is there, 8 is also there. so again suff.....

should be D.


We don't know this in Stmt 2! The source integer could either be 4, which would make 4, 16, and 64 in the set or 8, which would include 8, 64, 512

A is sufficient since we know 4 is not a source integer and can only be 2^2 and not a power of 3
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Re: A quick DS question [#permalink] New post 15 Aug 2007, 21:54
Fistail wrote:
amitavha wrote:
S is a set of positive integers such that if integer X is a member of S, then both X^2 and X^3 are also in S. If the only member of S that is neither the square nor the cube of another member of S is called the Source Integer, is 8 in S?

1) 4 is in S and is not the Source Integer.
2) 64 is in S and is not the Source Integer.

This is a question from Kaplan. Pls explain.

Thanks.



1: if 4 is in S and is not a source integer, 2 is there and subsequently 8 is there. so suff.........

2. if 64 is in set S, and 64 is not a source integer, 64 could have because of 4^3 or of 8^2. since 4 is there, 8 is also there. so again suff.....

should be D.


I will go with A.
In statement 2 we will be assuming that 4 is not a source integer. We know that 4 is 2^2 but can we assume this.
Going by the info that we have, I go with A.
Though i picked D initially thinking of 4^3 or 8^2.
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 [#permalink] New post 15 Aug 2007, 22:15
will go for D

the question states that -the only member of S that is neither the square nor the cube of another member of S is called the Source Integer.
to qualify as a non source integer would mean that the number is either the square of another number of the cube of another number .

As per statement 1- 4 is in S and is not the source integer. 4 is the square of 2 and since 2 is in the set we know there would be 8 ( as per statement is X is present then x^3 is also present) as well .
So sufficient .

As per statement 2- 64 is in S and its not the source integer.
64 is the square of 8 and the cube of 4 , so we know 8 is in the set.
Also sufficient .
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 [#permalink] New post 16 Aug 2007, 05:14
forgmat wrote:
will go for D

the question states that -the only member of S that is neither the square nor the cube of another member of S is called the Source Integer.
to qualify as a non source integer would mean that the number is either the square of another number of the cube of another number .

As per statement 1- 4 is in S and is not the source integer. 4 is the square of 2 and since 2 is in the set we know there would be 8 ( as per statement is X is present then x^3 is also present) as well .
So sufficient .

As per statement 2- 64 is in S and its not the source integer.
64 is the square of 8 and the cube of 4 , so we know 8 is in the set.
Also sufficient .


See red: That means 4 is source . So numbers in set should be 4 16 64 based on the question.Where is 8? 8 still could be there but in a different combination . { 8, 64 , 8^3 } . So there is two combination and B is not SUFF.
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 [#permalink] New post 16 Aug 2007, 08:20
yea see my mistake , i assumed 2 to be in the set, which is wrong . as 2 could be in the set or may not be .

thanks for the clatification.

answer should be A
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 [#permalink] New post 16 Aug 2007, 08:42
wow. tricky question to wrap your head around. I'm leaning towards A myself.
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 [#permalink] New post 16 Aug 2007, 19:10
Yeah, its the wording that confused me. Anyway, the answer is A. Thanks all for your input.

regds,
Amit
  [#permalink] 16 Aug 2007, 19:10
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