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S is a set of positive integers such that if integer X is a [#permalink]

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15 Aug 2007, 20:13

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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.

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.

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.

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.

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.

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.

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.

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 .

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.