Author 
Message 
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262
Location: Madrid

The numbers w,x,y and z are different integers. Is their sum [#permalink]
Show Tags
27 Jun 2007, 13:48
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please repost it in the respective forum.
The numbers w,x,y and z are different integers. Is their sum a multiple of 12?
(I) w,x,y and z are consecutive integers.
(II) z is a multiple of 24.



Manager
Joined: 14 Mar 2007
Posts: 235

i) insuff.> could be any four consecutive integers
ii) insuff> 24+any three integres does not guarantee that it will be divisible by 12
i+ii together if z is 24, so w is 21, x is 22 and y is 23. Their sum is 90 , that is not divisible by 12...and so on for z multiple of 24 and the other ones been consecutive
So answer for me is NO, and C



Manager
Joined: 15 Sep 2006
Posts: 94

Re: DS: Multiples [#permalink]
Show Tags
27 Jun 2007, 13:59
12= 3*2*2
Statement I.
Not sufficient. 1+2+3+4=10
Statement II.
Not sufficient.
If z=24, then 1+2+3+24=30, which is not a multiple of 12.
Combined:
Not sufficient.
My pick is E.



VP
Joined: 10 Jun 2007
Posts: 1438

Re: DS: Multiples [#permalink]
Show Tags
27 Jun 2007, 15:44
kevincan wrote: The numbers w,x,y and z are different integers. Is their sum a multiple of 12?
(I) w,x,y and z are consecutive integers. (II) z is a multiple of 24.
I got A
(I) 1+2+3+4 = 10, No.
Since the integers are consecutive, Sum = ((w+z) / 2 ) * 4 = 2 (w+z)
So now, we are finding: Does (w+z) ever a multiple of 6?
Since z = w+1+1+1 = w+3, w+z = 2w + 3. Since 3 isn't divisible by 6, this concludes that (w+z) will never be divisible by 6.
Because the sum will never be a multiple of 12, SUFFICIENT.
(2) z = 24 * (integer)
don't know anything about w,x,y
INSUFFICIENT.



Senior Manager
Joined: 21 Jun 2006
Posts: 282

I believe it is A.
Consecutive integers  w,x,y,z
take 1,2,3,4  sum is 10
2,3,4,5  14
3,4,5,6  18
sum is going to increase by 4 for each increase in digit
10,14,18,22,26,30,34,38....
Is the sum multiple of 12 ie, 12,24,36,48,60....
The two sets will never intersect, hence answer is no
SUFFICIENT
B is clearly insufficient
z=24, w=1,x=2,y=3,sum =30 not a multiple
z=24,w=12,x=36,y=48, sum = 120 which is a multiple
Hence insufficent.



Senior Manager
Joined: 04 Jun 2007
Posts: 345

Re: DS: Multiples [#permalink]
Show Tags
27 Jun 2007, 21:50
kevincan wrote: The numbers w,x,y and z are different integers. Is their sum a multiple of 12?
(I) w,x,y and z are consecutive integers. (II) z is a multiple of 24.
Doing it without putting numbers..
Let the four consecutive integers be n, (n+1), (n+2), (n+3)
Therir sum = 4n + 6 = 2*(2n+3)
So, the sum to be divisible by 12 (2n+3) must be divisible by 6. which is not possible for integers. Hence, the sum of four consecutive integers can never be divisible by 12.
Stmt 2 is clearly insufficient.



Director
Joined: 14 Jan 2007
Posts: 774

Stmt1:
let say numbers are
n, n+1, n+2, n+3
sum = 4n+6 =2(2n+3)
2n+3 is NOT divisible by 2 as this is an ODD number. Hence sum of 4 consecutive numbers will have exactly one factor of 2.
To be multiple of 12, we need more than one factor of 2.
So SUFF.
stmt2 is clearly insuff.
Hence answer should be 'A'










