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by francoisph » Tue Jun 01, 2010 5:53 am
Hi,

Please someone could explain me?

S is a set of positive integers such that if integer x is a member of S, the both x^2 and x^3 are also
in S. If the 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

thks
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by asamaverick » Tue Jun 01, 2010 6:22 am
1) 4 is in S and not a source integer. If a number is a source integer then its square and cube should be present in S, but the number itself need not be square or cube of another number. Since 4 is not a source integer, we can interpret that 2 is also present in S. 4 is in the set since it a square of 2. If 2 is present in S, then the cube of 2 (8) should also be present. Hence (1) is sufficient.

2) 64 is in S and not a source integer. Since 64 is not a source integer, it can be in the set as a square (of 8) or as a cube (of 4). If it is present in the set as a cube of 4 (and we do not know just by (2) alone if 4 is a source number or not) we cannot say if 8 is present in S. Hence (2) alone is not sufficient.

So think the answer should be A. What is the OA?
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by gmatmachoman » Tue Jun 01, 2010 7:48 am
asamaverick wrote:1) 4 is in S and not a source integer. If a number is a source integer then its square and cube should be present in S, but the number itself need not be square or cube of another number. Since 4 is not a source integer, we can interpret that 2 is also present in S. 4 is in the set since it a square of 2. If 2 is present in S, then the cube of 2 (8) should also be present. Hence (1) is sufficient.

2) 64 is in S and not a source integer. Since 64 is not a source integer, it can be in the set as a square (of 8) or as a cube (of 4). If it is present in the set as a cube of 4 (and we do not know just by (2) alone if 4 is a source number or not) we cannot say if 8 is present in S. Hence (2) alone is not sufficient.

So think the answer should be A. What is the OA?
sT 2 IS REALLY TRICKY.
AS rightly said, 64 is in the set. Now it's prescence is becox of 8 or 4 is a mystery.

Case 1 : 64 is in S. so is 8

YES

case 2: 64 is in S. So is 4.

What if 4 is Not source integer: The set will contain 2 & thereby 8/

But what if 4 is source integer :? 2 wont be there in set. So No 2 & No 8

In consistent. Insufficient!

Pick A
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by francoisph » Wed Jun 02, 2010 6:09 am
kaplan said

1 sufficient

2 sufficient
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by gmatmachoman » Wed Jun 02, 2010 9:50 am
francoisph wrote:kaplan said

1 sufficient

2 sufficient
Can u post the explanation of how St 2 is sufficient??
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by francoisph » Wed Jun 02, 2010 11:57 am
Statement (1): Sufficient
If 4 is in S, and it is not the source integer, then it must be the square
or cube of the source integer. Since 4 is not the cube of any positive
integer, and it is the square of 2, 2 must be in S and must be the source
integer. It follows that if 2 is the source integer, 4 is the square of 2,
and 8 is the cube of 2. So 2, 4, and 8 would be in S. Statement (1) by
itself is sufficient to answer the question.

Statement (2): Sufficient
If 64 is in S and it is not the source integer, it could represent the
square of 8 or the cube of 4. If it represents the square of 8, then 8, 64,
and 512 are in S. The number 8 would be a member of S in this case as well.

If 64 represents the cube of 4, then 4, 16, and 64 would be in S. Since 4
is the square of 2, 2 would be in S, and 8 (the cube of 2) would also be in
S.

Either statement by itself is sufficient to answer the question.
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by francoisph » Thu Jun 03, 2010 12:10 pm
KAPLAN S MISTAKE

it should be 1) sufficiency
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