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manhattan review word problem number properties

by resilient » Wed Mar 05, 2008 3:06 pm
If Z is an infinite subset of real numbers, is there a number in Z that is greater than every other number in Z?

1. Every number in Z is divisible by 5.

2.Every number in Z is a negative multiple of a prime number

qa is b

I dont undestand the question. any pointers!?
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Re: manhattan review word problem number properties

by Stuart@KaplanGMAT » Wed Mar 05, 2008 5:16 pm
Enginpasa1 wrote:If Z is an infinite subset of real numbers, is there a number in Z that is greater than every other number in Z?

1. Every number in Z is divisible by 5.

2.Every number in Z is a negative multiple of a prime number

qa is b

I dont undestand the question. any pointers!?
If Z is an infinite set, that means that it goes on forever. For one member in Z to be bigger than all the others, there would have to be a hard upper cap on the set.

For example, if Z were the set of all numbers less than or equal to 3, it would be infinite, but 3 would be bigger than all the other numbers in the set. On the other hand, if Z were the set of all multiples of 2, Z would continue upwards to infinity and there would be no "biggest number".

(1) Every number in Z is divisible by 5 - or, in other words, every number in Z is a multiple of 5.

Is it possible that there's a biggest number? Sure - (1) doesn't say that Z is EVERY multiple of 5, just that every number in the set is a multiple of 5 - Z could include every multiple of 5 up to 10000 (and down to negative infinity), in which case 10000 would be the biggest number. Is it possible that there isn't a biggest number? Also sure - Z could include every multiple of 5: insufficient.

(2) Every number in Z is a negative multiple of a prime number.

Well, we know that all prime numbers are positive. So, if every number in Z is a negative multiple of a prime, then every member of Z is negative.

So, the biggest possible member of Z is -2. Do we know for sure if -2 is in the set? No, but we know there's an upward boundary on the set. So, set Z will DEFINITELY contain a "biggest number" (even if we have no clue what that biggest number is, we know it exists): sufficient.

(1) is insufficient, (2) is sufficient: choose (b).
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by gmatguy16 » Wed Mar 05, 2008 6:32 pm
stuart, statement b does not say anywhere specifically that we cannot have repeated numbers...isnt it true that we can have 2 numbers as -2 in which case b is not sufficient?
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by hemanth28 » Wed Mar 05, 2008 7:42 pm
Stuart,

The question is if at all there is one number that is greatest and it's not which is the greatest number. --the answer should be 'yes' or 'no'
In any set there will always be a greatest number.And the greatest number is unique,however the element in the set may not be.
So according to me the answer is D.

Let me know if my understanding is wrong.
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by Stuart@KaplanGMAT » Wed Mar 05, 2008 8:00 pm
hemanth28 wrote:Stuart,

The question is if at all there is one number that is greatest and it's not which is the greatest number. --the answer should be 'yes' or 'no'
In any set there will always be a greatest number.And the greatest number is unique,however the element in the set may not be.
So according to me the answer is D.

Let me know if my understanding is wrong.
Infiinte sets don't necessarily have a biggest number - for example, the set of integers.
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by Stuart@KaplanGMAT » Wed Mar 05, 2008 8:19 pm
gmatguy16 wrote:stuart, statement b does not say anywhere specifically that we cannot have repeated numbers...isnt it true that we can have 2 numbers as -2 in which case b is not sufficient?
That, my friend, is a most excellent point!

One could probably make a technical argument that "number" doesn't mean the same thing as "term" - so that even if there are 17 -2s in the set, the "number" -2 is bigger than every other number. However, that's a pretty semantical argument and likely not going to be an issue on the real GMAT.

Sooo, looks like answer should probably be (e).

Any Manhattan GMATers have the official explanation for why we don't have to worry about identical terms?
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ANSWER IS B

by slicknstylish » Tue Mar 24, 2009 11:43 am
Z is a subset of R ( the real number set) .

R does not contain repetitive numbers (It only contains infinitely dense continuum of distinct real numbers, representing distinct points in the continuum).

Therefore, Z also does not contain repetitive numbers and -2 is the greatest element in Z, the answer is b.
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Re: ANSWER IS B

by Ian Stewart » Tue Mar 24, 2009 4:46 pm
Just to clear up a few points from the above:
slicknstylish wrote: Therefore, Z also does not contain repetitive numbers and -2 is the greatest element in Z, the answer is b.
As Stuart mentions above, there is no way to be sure whether -2 is even in the set from Statement 2 alone. We can be sure there is a largest element, but we can't be sure what that largest element is. Statement 2 really only tells us that every element in the set is a negative integer, and is less than -1, since every negative integer less than -1 is a negative multiple of a prime.
Stuart Kovinsky wrote: Any Manhattan GMATers have the official explanation for why we don't have to worry about identical terms?
I'm not sure that it's a Manhattan GMAT question - the thread title says it's from Manhattan Review, which is, of course, a different company altogether.
gmatguy16 wrote:stuart, statement b does not say anywhere specifically that we cannot have repeated numbers...isnt it true that we can have 2 numbers as -2 in which case b is not sufficient?
A real GMAT question would be less ambiguous on this point, since so many of the sets in real GMAT questions are what are properly known as 'data sets', and can contain repeated elements. In proper mathematical set theory, sets do not contain repeated elements. That's clearly the intended meaning in this question, though if it were a real GMAT question, I'm certain it would be phrased differently - the question would likely say that all the elements were 'distinct', for example.
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by dtwea » Wed Mar 25, 2009 1:05 pm
Since Z is a subset, I would find the Universal or Power Set. Let this be U.

1) says that Z is a multiple of 5. So what Universal set has as its elements only multiples of five.

U= { …… -15, -10, -5, 0, 5, 10, 15, …..} Since this is the universal set and Z is infinite, any combination from this set can be Z. Both U and Z have no limit.

Examples of Z: {0} (0, 5} {0, -5}, {U} { 15, -10, -5, 0, 5, 10, 15} and so on.

Now the question says, Is There a number in Z that is greater than every other number?
If Z is {0), the answer is No since the only number in Z is 0 and 0 cannot be greater than itself. If Z is {0, -5} the answer is Yes, since -5 is the only number in Z. That is all you need to rule out (1). It does not matter which Z you take as long as Z has only multiples of 5.

b). Remember Z is a subset. We know that X is a multiple of Y if X = MY, where X, Y and M are all integers. So the elements in Z are negative multiples of prime numbers, which is a fancy way of saying the elements in Z are negative prime numbers! Our universal set U is the set of prime numbers on number line and we can construct Z from this set.

Z = {…… -17, -13, -11, -7, -5, -3, -2, -1, }. Remember any combination of these as a subset would do. { -17, -13}, {-3,-1) and so on.

Now is there a number from Z, any Z, that is greater than every other number in Z stretched to infinity? The answer is Yes. It is -1.
B is Correct.

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by Ian Stewart » Wed Mar 25, 2009 1:49 pm
dtwea wrote:
Examples of Z: {0} (0, 5} {0, -5}, {U} { 15, -10, -5, 0, 5, 10, 15} and so on.
Z is an infinite subset, according to the question, so it must contain an infinite number of elements. Z cannot be {0}, for example.
dtwea wrote: b). Remember Z is a subset. We know that X is a multiple of Y if X = MY, where X, Y and M are all integers. So the elements in Z are negative multiples of prime numbers, which is a fancy way of saying the elements in Z are negative prime numbers!
That's not quite the case. A number like -24 is a "negative multiple of a prime", since it's equal to 3*(-8). There are potentially other numbers to consider here besides the negative equivalents of the prime numbers.
dtwea wrote: Z = {…… -17, -13, -11, -7, -5, -3, -2, -1, }. Remember any combination of these as a subset would do. { -17, -13}, {-3,-1) and so on.
Note that 1 is not a prime number, so -1 is not a negative multiple of a prime. The largest element that could possibly be in our set, using Statement 2, is -2.
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by aj5105 » Sun May 03, 2009 10:49 pm
Good point !
gmatguy16 wrote:stuart, statement b does not say anywhere specifically that we cannot have repeated numbers...isnt it true that we can have 2 numbers as -2 in which case b is not sufficient?
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