Zero is a even integer

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Zero is a even integer

by leyley1390 » Sat Feb 23, 2013 8:50 am
Hi,
I had this question on the GMAT:

if x,y,z are three integers, are they consecutive integers?

(1) z=x+2
(2) None of the three are multiple of 3.

I would select E because 0,1,2 are three consecutive integers, and none of them are multiple of 3.
But my book said that B is the correct answer because in any set of three consecutive integers, on of the three is automaticlly multiple of 3.

Who is right, me or my book?

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by Anurag@Gurome » Sat Feb 23, 2013 9:01 am
leyley1390 wrote:I would select E because 0,1,2 are three consecutive integers, and none of them are multiple of 3.
0 is multiple of every integer.
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by Brent@GMATPrepNow » Sat Feb 23, 2013 9:03 am
leyley1390 wrote:Hi,
I had this question on the GMAT:

if x,y,z are three integers, are they consecutive integers?
(1) z=x+2
(2) None of the three are multiple of 3.
Target question: Are the three integers consecutive?

Statement 1: z=x+2
There are several sets of values that meet this condition. Here are two:
Case a: x=1, y=2, z=3, in which case the 3 integers are consecutive.
Case b: x=1, y=8, z=3, in which case the 3 integers are not consecutive.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: None of the three are multiple of 3.
There's a nice rule that says: If there are n consecutive integers, then exactly 1 of them is divisible by n
Example: Of the 5 consecutive integers 7, 8, 9, 10, 11, exactly one is divisible by 5 (or we can say one is a multiple of 5.
Since, none of the 3 integers is divisible by 3, then the 3 integers must not be consecutive.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

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Brent
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by leyley1390 » Sat Feb 23, 2013 9:08 am
Ok thanks

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by The Iceman » Sat Feb 23, 2013 10:00 am
Brent@GMATPrepNow wrote: There's a nice rule that says: If there are n consecutive integers, then exactly 1 of them is divisible by n
Exactly! For the same reason the product of k consecutive integers is perfectly divisible by k! (factorial of k)