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GMATPrep Integer Problem

by Guest Thu Nov 13, 2008 2:55 pm

I am having problems understanding why the correct answer for the following problem is correct.


Is the integer x divisible by 6?
(1) x + 3 is divisible by 3
(2) x + 3 is an odd number.

Answer choices:
(a) Statement 1 alone is sufficient, but Statement 2 alone is not sufficient.
(b) Statement 2 alone is sufficient, but Statement 1 alone is not sufficient.
(c) Either statements are NOT sufficient, but BOTH together are sufficient.
(d) Each statement ALONE is sufficient.
(e) Neither statement Alone is sufficient.

Correct answer: c

Any one have any insights?
RR
 
 

by RR Fri Nov 14, 2008 9:26 am

1. x + 3 is divisible by 3
This means that x is a multiple of 3. But multiples of 3 can be 3, 6, 9, 12,...
Note that the multiples of 3 which are odd numbers like 3, 6 etc are not divisible by 6
Insufficient. Eliminate A, D

2. x + 3 is an odd number
If x + 3 is odd, then x can be any even number like 4, 6, 10 etc.
Insufficient. Eliminate B, D

1 & 2 . x is a multiple of 3 and x is even ie x can be 6, 12, 18 etc. Sufficient. Answer is C
RonPurewal
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by RonPurewal Thu Nov 20, 2008 8:02 am

RR Wrote:1. x + 3 is divisible by 3
This means that x is a multiple of 3. But multiples of 3 can be 3, 6, 9, 12,...
Note that the multiples of 3 which are odd numbers like 3, 6 etc are not divisible by 6
Insufficient. Eliminate A, D

2. x + 3 is an odd number
If x + 3 is odd, then x can be any even number like 4, 6, 10 etc.
Insufficient. Eliminate B, D

1 & 2 . x is a multiple of 3 and x is even ie x can be 6, 12, 18 etc. Sufficient. Answer is C


this is a good solution.

here's a pertinent takeaway.
TAKEAWAY:
problems that deal with REMAINDERS
(usually, but not always, in conjunction with divisibility) tend to involve patterns that REPEAT in easily identifiable ways.
these patterns also tend to come out of the woodwork fairly quickly, so you can solve lots of these problems by just grinding out sample numbers, as the user "rr" above has done, until the pattern becomes obvious.
LP1
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Re: GMATPrep Integer Problem

by LP1 Fri Apr 24, 2009 7:23 pm

hi guys,
Since 0 is also a possible value of x (no constraints in Question stem stating otherwise) and since 0 is considered an even integer, why is it not considered as a possible value of x making choice E correct. Ofcourse since C is the correct choice 0 is not an option but I was wondering why not?

thanx
RonPurewal
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Joined: Tue Aug 14, 2007 8:23 am
 

Re: GMATPrep Integer Problem

by RonPurewal Mon Apr 27, 2009 4:20 am

LP1 Wrote:hi guys,
Since 0 is also a possible value of x (no constraints in Question stem stating otherwise) and since 0 is considered an even integer, why is it not considered as a possible value of x making choice E correct. Ofcourse since C is the correct choice 0 is not an option but I was wondering why not?

thanx


well, technically, 0 is a multiple of 6, so this is still (c).

18 is a multiple of 6, because 18 is 3 times 6, and 3 is an integer.
30 is a multiple of 6, because 30 is 5 times 6, and 5 is an integer.
0 is a multiple of 6, because 0 is 0 times 6, and 0 is an integer.

--

i've never known the gmat to test 0 as a multiple of things, though. (it's actually a multiple of EVERY integer.)
i have a strong suspicion that this problem statement was supposed to contain "POSITIVE integer", and someone left that off.

this suspicion is intensified by the somewhat sloppy copying of the answer choices (like c: "either statements are..."). wherever this problem came from, it's certainly not verbatim.

TO THE ORIGINAL POSTER:
do you have a screen shot?