Is zero divisible by any integer?

[solliday]

I noticed in the second lab that on one of the questions he doesn't list 0 as a possible integer divisible by 9 or 12, although I thought that zero was divisible by any number. What's the rule?

[RonPurewal]

hi.

the full disclosure is:
* 0 is divisible by any positive integer
BUT
* this won't matter on the test.

the gmat is not going to test divisibility, other than for positive integers. i.e., they will not test you on the divisibility properties of 0, or on those of negative numbers.

the gmat is VERY "nice" about not testing these sorts of nit-picky aspects of definitions.
for instance:
* they're not going to test the fact that 1 is not prime, even though they technically could.
* they're not going to use the phrase "the sum of all the numbers in the set" when the set consists of one number, even though they technically could.
etc.

although they certainly write problems that are devious in other ways, they will NOT write problems that turn on the unexpected minutiae of definitions.

hope that reassures you.

[evanm223]

To bring up an old topic: I'm seeing something on one of the GMATPrep (official) exams.

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

My answer: E
Official answer: C


(1) x=0; ????????
x=3; NO
x=6; YES
INSUFFICIENT

(2) i.e. x is even
x=0; ?????
x=2; NO
x=6; YES
INSUFFICIENT

(1+2) cases x=0 and 6 both hold.

If anyone could please help me shed light on this, I'd appreciate it. And in case it crosses your mind: no, it doesn't say x is a POSITIVE integer; it just says "integer".

Does this mean the GMAT does, in fact, test divisibility rules for 0?

Thanks in advance.

[RonPurewal]

well, 0 is in fact divisible by every integer.

if you apply any valid definition of "divisible", then zero will pass the test.

e.g., " 'm is divisible by n' means that m/n is an integer"
okay, 0/6 = 0 = integer.

or, " 'm is divisible by n' means that m is an integer multiple of n"
okay, 0 is equal to zero times 6, so it's an integer multiple of 6.

[RonPurewal]

you do have a point, though. this problem is in the software, but, if it were put onto the official exams as an experimental problem, i don't think it would pass the test.

in other words, i don't think you will ever see anything like this on an OFFICIAL exam.

[RonPurewal]

note also that, if you just ignore the "weird" case, you get the correct answer.

this is not a coincidence; they don't want to create problems that rely on tricky quirks.

so, perhaps this is the best advice:
if you come across a "weird" case, pick the answer you'd get by IGNORING that case.

[JavierC840]

hi guys, I do not understand your discussion. To me the answer is C with or without considering x=0.

from "x+3 is divisible by 3" the result is that x must be one of the following: ...-12, -9, -6 -3, 0, 3, 6, 9,12...
some are divisible by 6 and some are not: INSUFFICIENT

from "x+3 is odd" the result is that x must be even, i.e. x must be one of the following: ...-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 12...
some are divisible by 6 and some are not: INSUFFICIENT

if you consider both conditions together it means that x must be any number that is in both of the sets above. These numbers are: ...-12, -6, 0, 6, 12...and these are all divisible by 6 (including x=0). Therefore, the correct answer is C and the question about the divisibility of zero should not be an issue to it.

[RonPurewal]

hi guys, I do not understand your discussion. To me the answer is C with or without considering x=0.

this ^^ is exactly the point of the post directly above yours: if you ignore zero (and negatives), the result is the same.