Integer set problem

This topic has expert replies
User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Thu Feb 03, 2011 11:07 am
Location: Chennai, India
Thanked: 1 times
Followed by:1 members

Integer set problem

by Vishnu88 » Wed Apr 13, 2011 7:50 am
S is a set of positive integers such that if integer X is a member of S, then both X2 and X3 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

Question is from KAPLAN Math workbook.

User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 am
Thanked: 72 times
Followed by:6 members

by manpsingh87 » Wed Apr 13, 2011 8:04 am
Vishnu88 wrote:S is a set of positive integers such that if integer X is a member of S, then both X2 and X3 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

Question is from KAPLAN Math workbook.
1) if 4 is in s , then it means, 16 and 64 are also in s, also 4 is not a source integer, then it means it must be either a square or cube of another integer, we know that sqrt(4)=2 or -2, as we are dealing with only positive numbers hence sqrt(4)=2, therefore 2 is also part of s, if 2 is there than 8 will also be there in s, therefore 1 alone is sufficient to answer the question.

2)64 is in s, than it means 64^2 and 64^3 will also be in s, also 64 is not a source integer, therefore either its square root will also be in s, i.e. sqrt(64)=8 can be in s or cuberoot(64)=4 can also be in s, hence 2 alone is not sufficient to answer the question, hence A
Last edited by manpsingh87 on Fri Apr 15, 2011 7:07 am, edited 3 times in total.
O Excellence... my search for you is on... you can be far.. but not beyond my reach!

Master | Next Rank: 500 Posts
Posts: 423
Joined: Fri Jun 11, 2010 7:59 am
Location: Seattle, WA
Thanked: 86 times
Followed by:2 members

by srcc25anu » Fri Apr 15, 2011 5:58 am
stat 1: 4 is in S and is not the source integer
therefore numbers in S can be 4 16 64 (8 can be in S) or 2 4 8 (8 cannot be in S) hence not sufficient

stat 2: 64 is in S and is not the source integer
therefore numbers can be 8 64 64^2 (8 cannot be in S) or 4 16 64 (8 can be in S) hence not sufficient

Together: S can have 4 16 64 plus the source integer which can be either 8 or 3,5,7... etc
hence both statements together are not sufficient.
IMO: E

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Fri Apr 15, 2011 7:05 am
S is a set of positive integers such that if integer X is a member of S, then both X2 and X3 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.
Only the source integer is neither the square nor the cube of another member of S.
Thus, if X is not the source integer, then X is either the square or the cube of another member of S.

Statement 1: 4 is in S and is not the source integer.
Thus, 4 is either the square or the cube of another member of S.
Since 4 is not the cube of an integer, 4 must be the square of another member of S.
Thus, 2 is in S.
If 2 is in S, then 2²=4 and 2³=8 are in S.
Sufficient.

Statement 2: 64 is in S and is not the source integer.
Thus, 64 is either the square or the cube of another member of S.
If 64 is the square of another member of S, then 8 is in S.
If 64 is not the square but only the cube of another member of S, then we know that 4, 4²=16 and 4³=64 are all in S, but we cannot determine whether 8 is in S.
Insufficient.

The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3