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DS: Set of integers: Princeton review math bin 3 question

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by dinesh19aug » Sun Jul 25, 2010 1:23 pm
Q9: Q9: If P is a set of integers and 3 is in P, is every positive multiple of 3 in P?
1) For any integer in P, the sum of 3 and that integer is also in P.
2) For any integer in P, that integer minus 3 is also in P.

My Approach:
I picked statement 2 first: 2) x-3 is always present in P, so I pick X up 1,2,3, 0, -1 ===>
x-3 will be ==> -2 , -1, 0, 3, -4 Not sufficient.

Statement 2) Same as 1. Not sufficient

Both statement NOT SUFFICIENT

The OA is D.
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Source: — Data Sufficiency |
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by dinesh19aug » Sun Jul 25, 2010 1:27 pm
Actually now that I just posted this question. It made me think again and I know what I did wrong. I CANNOT pick a random number.

The first number that I should pick is 3, as it is given in question that 3 is one f the number in P.

Hence Stmt 2: 3-3 = 0
0-3 = -3
-3 -3 = -6 .... This gives all the negative multiples of 3.

Stmt 1: 3 + 3 = 6
6+3 = 9 .... gives all the positive multiple of 3.

Hence both stmt are good enough to find the solution independently .... So the answer is D
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by outreach » Mon Jul 26, 2010 12:05 am
answer shd be A

Positive multiples of 3 are: 3, 6, 9, 12, 15,

stmt 1

if x is in 3, then x+3 is also in the set
already 3 is in set so 3+3 will also be in the set..values of set will be (3,6,9.....)
elements in the set r positive multiples of 3
suff

stmt 2
if x is in 3, then x-3 is also in the set
already 3 is in set so 3-3 will also be in the set..values of set will be (3,0,-3,-6.....)
all elements in the set r not positive multiples of 3

insuff
dinesh19aug wrote:Actually now that I just posted this question. It made me think again and I know what I did wrong. I CANNOT pick a random number.

The first number that I should pick is 3, as it is given in question that 3 is one f the number in P.

Hence Stmt 2: 3-3 = 0
0-3 = -3
-3 -3 = -6 .... This gives all the negative multiples of 3.

Stmt 1: 3 + 3 = 6
6+3 = 9 .... gives all the positive multiple of 3.

Hence both stmt are good enough to find the solution independently .... So the answer is D
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by marclife » Thu Mar 21, 2013 5:37 am
the question doesn't ask if every number in the set is positive. It only asks if every positive multiple of the is in the set. If I test any integer greater than 6 (6,9,12,15,18) I get positive multiples of 3. It seems like a poorly written question.
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by srcc25anu » Thu Mar 21, 2013 2:23 pm
we know that set P contains 3 (given). now we need to find out whether all positive multiples of 3 are in P or not?
lets denote that integer in P is INT

A states sum of 3 + INT must be in P. we know 3 is in P so 3 + 3 = 6 is in P, 6+3 = 9 is there, similarly 12, 15, 18 .... infinity all +ve integers are there in P
also if 3 = INT + 3, then INT = 0, so 0 is present in set P, if 0 = INT + 3, then INT = -3 is also in set P.and likewise 0, -3,-6,-9 ... negative infinity are all present in P.

Are all positive multiples of 3 present in P? YES. This is sufficient condition.

B states 3 - integer in P is also present in P. going by the same logic above, we know 3 is present in P. lets say that integer in P = INT.
so if 3 - 3 = 0 present in P
0 - 3 = -3 present in P
-3 - 3 = -6 present in P
likewise all negative multiples and 0 are included. in addition we have a few more elements in set P
since 3 is present, INT - 3 = 3 or INT = 6 is also present.
INT - 3 = 6 or INT = 9 is also present.
likewise, we also have all positive multiples of 3 in set P.

Again do we have all positive multiples of 3 in set p? YES. hence sufficient

this can be answered by both statements individually hance answer is D.
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by GMATGuruNY » Thu Mar 21, 2013 8:18 pm
dinesh19aug wrote:Q9: Q9: If P is a set of integers and 3 is in P, is every positive multiple of 3 in P?
1) For any integer in P, the sum of 3 and that integer is also in P.
2) For any integer in P, that integer minus 3 is also in P.
Statement 1: For any integer in P, the sum of 3 and that integer is also in P.
In other words, if we ADD 3 to any integer in P, we get ANOTHER INTEGER in P.
Since 3 is in P, 3+3=6 also is in P.
Since 6 is in P, 6+3=9 also is in P.
Since 9 is in P, 9+3=12 also is in P.
And so on.
Thus, every positive multiple of 3 -- {3, 6, 9, 12...} -- is in P.
SUFFICIENT.

Statement 2: For any integer in P, that integer minus 3 is also in P.
In other words, if we SUBTRACT 3 from any integer in P, we get ANOTHER INTEGER in P.
Since 3 is in P, 3-3=0 also is in P.
Since 0 is in P, 0-3=-3 also is in P.
Since -3 is in P, -3-3=-6 also is in P.
And so on.
Thus:
Every multiple of 3 LESS THAN OR EQUAL TO 3 -- {3, 0, -3, -6...} -- is in P.
No way to determine whether any multiples of 3 GREATER THAN 3 are in P.
INSUFFICIENT.

The correct answer is A.

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