Data Sufficiency

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!?

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).

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?

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.

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.

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?

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.

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.

Beautiful Question!!!

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?