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Help with very tough Number Properties problem.

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Help with very tough Number Properties problem.

by jeremy8 » Mon Jul 26, 2010 2:42 pm
K is a set of numbers such that

(i) if x is in K, then -x is in K, and
(ii) if each of x and y is in K, then xy is in K.

Is 12 in K?

1) 2 is in K
2) 3 is in K

OA C

I don't really understand what x and y stand for in this problem. The official guide's explanation (this is DS#70 from Quant Review 12th ed.) doesn't make any sense to me.
To me, 2 is in K stands for x is in K, and therefore, -2 is in K as well according to (i). That's it.
2) tells us that 3, and therefore -3 is in K. I assume 3 stands for y, so 1) and 2) together means that 2;-2;3;-3; and 6 (xy) are in K, end of story. So it's C. Both 1) and 2) are enough to say that 12 is NOT in K

However, the OG's explanation is that 12 is indeed in K. They assume from 1) that K could be the set of all real numbers or the set {..., -16, -8. -4, -2, 2, 4, 8, 16, ...}
I have no idea how they assume this from (i) and (ii). This makes no sense at all to me, please help!
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by Brian@VeritasPrep » Mon Jul 26, 2010 3:37 pm
Hey Jeremy,

Thanks for posting - I love this problem for many of the reasons that you mentioned. x and y seem fairly ambiguous, which is where the difficulty comes from, but once you catch this concept you should be able to apply it widely:

x just means "any number" - it's a variable. The same thing goes for y - it can mean any number.

statement i tells us that, for any number in this set, its negative is also in the set. If we combine i with statement 1, we'd see that:

2 is in K, which also means that its negative is in K, so the set includes (but is not limited to) 2 and -2.


statement ii tells us that, for any two numbers in this set, their product is also in the set. In other words, take any two numbers in this set and multiply them together, and that number is in the set. Combining these facts with statement 1 tells us that, if 2 is in the set, then we know that -2 is in the set (from that first point, i), which means that -4 is in the set, 8 is in the set, 16 is in the set, and so on.

Moving forward, though, statement 1 does not give us a multiple of 3, so we don't know whether 12 is in the set. Same for statement 2, which says that 3 is in the set (so therefore -3, -9, 27, etc.), but we don't have a multiple of 2.

Take them together, however, and we know that 2 and 3 are in the set, which means:

by virtue of i), -2 and -3 are in the set.
by virtue of ii), then 2*3 = 6 is in the set, and if 6 is in the set then 2*6 is in the set, so 12 is, indeed, in set K.

What's probably most important here is knowing that x and y don't represent specific variables - they're there to show that these rules (i) negatives and ii) products) hold for any number in the set.

I hope that helps...
Brian Galvin
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Veritas Prep

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by sk818020 » Mon Jul 26, 2010 3:39 pm
First you should break down 12 into it's prime factors to see what sort of base numbers you will need to work with.

12=(2^2)(3). From the rules, if 12 is going to be in set K, it will need at least one 2 and one 3. Because with that you will also have -2 and -3. Thus, 2*-2*-3=12 or if 2 is in the set then -2 will be in the set and so will -4 (2*-2) and if you have a 3 you will also have -3 and -4*-3=12.

Lets look at the answers.


1) 2 is in K.

If 2 is in K, then by (i) -2 is in K. If both 2 and -2 are in K, then by rule (ii) -4 must be in K also (2*-2=-4). If -4 is in K then, 4 (i) is in K. You can also conclude that any multiple of 2, -2, 4 and -4 must also be in K. 1) only tells us that multiples of two are in the set.

There is no mention of a 3 so we can rule this out. Insufficient.

2) 3 is in K.

From this we can conclude that -3 is in the set from rule (i), -9 by rule (iI), 9 by rule (i)... and so on.

This does not give us any indication of a 2 being in the set, so 2) is insufficient.

You combine them and you know that there is a 2 and a 3 in the set and by the reasoning at the beginning of the explanation, you will have have 12 in the set.

C.

Hope this helps.

Thanks ,

Jared
Last edited by sk818020 on Mon Jul 26, 2010 3:45 pm, edited 1 time in total.
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by jeremy8 » Mon Jul 26, 2010 3:43 pm
Brian, Jared,

Thanks a lot for your answers.

I was definitely confused about the "language" of the problem and what I could legitimately assume from the statements, but it's much clearer now and it finally makes sense.
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