Set of Integer Question

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smclean23: Master | Next Rank: 500 Posts; Posts: 172; Joined: Mon Jun 23, 2008 12:00 pmFollowed by: 1 member; Thanked: 3 times

by smclean23 » Mon Jul 07, 2008 4:08 pm

S is a set of integers such that
i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is –4 in S?

(1) 1 is in S.
(2) 2 is in S.

Answer is B

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ildude02: Master | Next Rank: 500 Posts; Posts: 320; Joined: Sun Jan 13, 2008 10:00 pm; Thanked: 10 times

by ildude02 » Tue Jul 08, 2008 8:58 am

You can deduce the question to, S(a) = {a, -a} and also S(a, b)= {a, b, ab}

Now from statement 1, we know S has 1 and -1. But at the same time, may be S also has 4, means -4 is in the set, so INSUFFICIENT.

From 2, S={2, -2} and since we can treat a= 2, b =-2, ab = -4 is also in set S, so SUFFICENT.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Fri Jul 11, 2008 2:45 am

You can deduce the question to, S(a) = {a, -a} and also S(a, b)= {a, b, ab}

i love this sort of notation - it's exciting, in its own cute little way.
two things:
(1) the second of these is unnecessarily limited; the set generated by a and b includes not only a, b, and ab, but also ab^2, (a^2)(b^3), and so on - in fact, any combination of a's, b's, and negative signs.

(2) i prefer the following intuitive explanations:
"if a is in S, then –a is in S" --> this means that you can flip the sign of a number and still stay in the set.
"if each of a and b is in S, then ab is in S" --> this means that you can multiply two numbers in the set, or one number times itself, and stay in the set.

so basically, you're trying to see if you can arrive at -4 by following these two types of directions.

statement (1):
if 1 is in S, then we can flip the sign and find that -1 is also in S.
but 1 x 1 = 1, so the multiplication rule isn't going to generate any new numbers. therefore, the only members that are guaranteed to be in S are 1 and -1.
insufficient.
NOTE: this is NOT to say that 1 and -1 are the only numbers in S. S could contain a whole congeries of other numbers, which is why the statement is insufficient. (if we actually knew that the set was limited to 1 and -1, this statement would actually be sufficient.)

statement (2):
if 2 is in S, then we can flip the sign and find that -2 is also in S. multiplying these two numbers gives 2 x -2 = -4, so -4 is in S.
sufficient.

answer = b

--

here are a couple of cool variations on this problem:
* change the question to "is -32 in S?" (answer is still b)
* change statement 2 to "16 is NOT in S" (answer is still b, although for different reasons this time - post back if you can't figure out why)

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target790: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Sun May 04, 2008 12:18 pm; Thanked: 1 times

by target790 » Fri Jul 11, 2008 3:09 pm

Ron,let me double check with you w.r.t two twists you added in the post

1>Is -32 in S

considering statement 2, S will be
{2,-2,-4,4,-8,8,16,-16,32,-32.......}

So B is sufficient

2>16 is not in S

so 16,-16 is not there as well.In this scenario any factors of 16,-16 should not be there in S as well(say 2,-2,4,-4,-8,8 )

Let me know if my assumption is not correct.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Fri Jul 11, 2008 7:49 pm

target790 wrote: 1>Is -32 in S

considering statement 2, S will be
{2,-2,-4,4,-8,8,16,-16,32,-32.......}

So B is sufficient

correct.

target790 wrote: 2>16 is not in S

so 16,-16 is not there as well.In this scenario any factors of 16,-16 should not be there in S as well(say 2,-2,4,-4,-8,8 )

close, but not quite correct.

you can have some factors of 16 without actually being able to make 16. for instance, if we only know that S contains 8 and -8 (and not, say, +/- 2 or +/- 4), then we could only put those together to make +/- 64; we wouldn't be able to generate intermediate powers of 2 such as 16 or 32.
remember, you can't work backwards from a number. you can work forwards - e.g., if 3 is in the set, then so are +/- 9, 27, 81, etc. - but the inverse isn't true; if, say, 81 is missing from S, we don't know whether S contains 27 (although we do know that 3 and 9 are missing; see below).

the real issue lies in working forward from the presence of certain factors, and in using contrapositive reasoning (if P then Q --> if not-Q then not-P), or, if you prefer, reductio ad absurdum.
here's how you settle the second part:
if -4 WERE to be in S, then you could multiply it by itself to get 16, contradicting the assumption that 16 is not in the set.
therefore, -4 can't be in the set.
sufficient.

note that "16 is not in S" is insufficient to settle the question of whether 8 is in S, even though 8 is a factor of 16.

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target790: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Sun May 04, 2008 12:18 pm; Thanked: 1 times

by target790 » Sat Jul 12, 2008 2:04 am

thanks much for the explanation

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Aalriy: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed Aug 27, 2008 8:49 pm

by Aalriy » Mon Sep 08, 2008 3:17 am

Y is Stmt 1 not sufficient to answer the question??

U have 1 in S. i.e u also have -1--> i)
And u also have 1 * -1= -1 ... -->ii)

So, S ={1, -1}.
U can answer that -4 is not there in S. Y is it not sufficient. Shouldnt the answer be D.

Tell me if i am wrong.

impossible is notthing....

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Mon Sep 08, 2008 4:03 am

Aalriy wrote:Y is Stmt 1 not sufficient to answer the question??

U have 1 in S. i.e u also have -1--> i)
And u also have 1 * -1= -1 ... -->ii)

So, S ={1, -1}.
U can answer that -4 is not there in S. Y is it not sufficient. Shouldnt the answer be D.

Tell me if i am wrong.

Your calculations are correct- that is, if you know that 1 is in the set, you can be sure that -1 is in the set. So the set might be {1, -1}. But that doesn't mean that 1 and -1 are the only things in the set. The set might also be {..., -9, -7, -5, -3, -1, 1, 3, 5, 7, ...} or {...-3, -2, -1, 0, 1, 2, 3, ...}. In either case, for any a you choose, -a is in the set, and for any a and b you choose, ab is in the set.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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Aalriy: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed Aug 27, 2008 8:49 pm

by Aalriy » Mon Sep 08, 2008 9:19 pm

Ian Stewart wrote:
Aalriy wrote:Y is Stmt 1 not sufficient to answer the question??

U have 1 in S. i.e u also have -1--> i)
And u also have 1 * -1= -1 ... -->ii)

So, S ={1, -1}.
U can answer that -4 is not there in S. Y is it not sufficient. Shouldnt the answer be D.

Tell me if i am wrong.
Your calculations are correct- that is, if you know that 1 is in the set, you can be sure that -1 is in the set. So the set might be {1, -1}. But that doesn't mean that 1 and -1 are the only things in the set. The set might also be {..., -9, -7, -5, -3, -1, 1, 3, 5, 7, ...} or {...-3, -2, -1, 0, 1, 2, 3, ...}. In either case, for any a you choose, -a is in the set, and for any a and b you choose, ab is in the set.

Ian Thanx for the explanation.

impossible is notthing....

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dwilliams05: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Tue Jan 06, 2009 1:07 pm

still don't get it

by dwilliams05 » Fri Sep 18, 2009 6:11 pm

Lunar, I still dont understand how the ii) can be taken to mean two numbers multiplied together OR the number multiplied by itself? Isn't that an assumption or am i missing something in wording of this question. Sorry, I know this is a pretty old question

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crackthetest: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Mon Feb 16, 2009 8:50 pm; Thanked: 3 times

question

by crackthetest » Fri Sep 18, 2009 8:10 pm

Lunar, here the stmt (ii) specifically states each of a and b which means a <> b correct?

If so your argument of a^2 or b^3 may not be in the set.

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