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If S is a set of integers and 5 is in S, is every multiple [#permalink]
 17 Jan 2004, 23:04
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If S is a set of integers and 5 is in S, is every multiple of 5 in S?
1) if x is in S, then x+5 is in S
2) if x is in S, then x-5 is in S
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Intern
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C?
1: if 5 is in S, so is 10, so is 15 and so on...20,25 because x+5
not suff because it only goes up the positive line
2: if 5 is in S, so is 0, -5, -10 and so on because x-5
not suff because it only goes down the negative line
put then to together.
5,10,15...and 5, 0, -5, -10, -15...
and you get all multiples of 5
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i apologize, but in my previous posting i posted the answer was E,
the actual anser is C....anyone know why?
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Manager
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I dont think C is correct , let set={5,6,10,15,19,23,25,10000000001} and let x=10 then (x+5) 15 is in Set and (x-5) 5 is in set but the set doesnt contain all the multiples of 5.
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What is the source of this question? It's very unclear...
If S is a set of integers and 5 is in S, is every multiple of 5 in S?
1) if x is in S, then x+5 is in S
2) if x is in S, then x-5 is in S
This question could be interpreted in a number of ways....
First, I picked E. 1) Cause let's say that x is 5, then 5 and 10 will be in the set. Not sufficient. 2) Let's say that x is 5, then 5, and 0 will be in the set. Not sufficient. 3) Again, let's use 5.. then 5, 10, 0 will be in the set. Not sufficient. Also, you can use other numbers that are not mutliples of 5.
To accurately get the answer C, then the question needs to be reworded to something like this... For every member of set S (called x), then x+5 is in S. Or something to that "ETS verbage" stuff. The question as it is, implies that x is only one member of the set, and you use the +5 or -5 rule to only one member, not all the members.
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Senior Manager
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gmatanh wrote: What is the source of this question? It's very unclear...
If S is a set of integers and 5 is in S, is every multiple of 5 in S?
1) if x is in S, then x+5 is in S
2) if x is in S, then x-5 is in S
This question could be interpreted in a number of ways....
First, I picked E. 1) Cause let's say that x is 5, then 5 and 10 will be in the set. Not sufficient. 2) Let's say that x is 5, then 5, and 0 will be in the set. Not sufficient. 3) Again, let's use 5.. then 5, 10, 0 will be in the set. Not sufficient. Also, you can use other numbers that are not mutliples of 5.
To accurately get the answer C, then the question needs to be reworded to something like this... For every member of set S (called x), then x+5 is in S. Or something to that "ETS verbage" stuff. The question as it is, implies that x is only one member of the set, and you use the +5 or -5 rule to only one member, not all the members.
I am sorry but I do not agree with you. The given wording of the problem is accurate.
When the statement says that if x is in S, then x+5 is in S, then you are invited to think recursively.
We have already been given that 5 is in S, so based on the first statement 5, 10, 15, ..... infinite all are members of S. Similarly the second statment also apllies. The answer should be C.
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Dudes, my answer is A.
Cause the multiples of 5 are 5, 10, 15, ect. In other words, set of POSITIVE numbers only. The way of thinking stays the same. 
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Senior Manager
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Solo wrote: Dudes, my answer is A. Cause the multiples of 5 are 5, 10, 15, ect. In other words, set of POSITIVE numbers only. The way of thinking stays the same. I disagree. The multiple of any integer includes 0, -ve integers and +ve integers.
If you are interested in more sofisticated defination, here it is.
The set of all multiples of an integer a will be denoted by
{ m in Z | m = aq for some q in Z }. where Z is a set of integers {....-3, -2, -1, 0, 1, 2, 3,......}
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It's news to me that 0 is a multiple of 5
This was not the case where I went to school
Though this expains the answer to one of the Kaplin
maths probems in the 2004 GMAT book
Question 13 in the maths section practice test states that 0 is a multiple
of 5
I've just got to get out more 
_________________
Don't die wondering
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Quote: This was not the case where I went to school
Were you taught by a pack of wolves?
http://mathforum.org/library/drmath/view/60913.html
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That was too snarky by half. Sorry.
The stress of my looming test is getting to me.
calnhob's answer is spot-on, IMO.
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close
