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The numbers w,x,y and z are different integers. Is their sum

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The numbers w,x,y and z are different integers. Is their sum [#permalink] New post 27 Jun 2007, 12:48
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B
C
D
E

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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.
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 [#permalink] New post 27 Jun 2007, 12:58
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
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Re: DS: Multiples [#permalink] New post 27 Jun 2007, 12:59
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.
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Re: DS: Multiples [#permalink] New post 27 Jun 2007, 14:44
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.
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 [#permalink] New post 27 Jun 2007, 20:11
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.
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Re: DS: Multiples [#permalink] New post 27 Jun 2007, 20:50
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.
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 [#permalink] New post 28 Jun 2007, 01:56
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'
  [#permalink] 28 Jun 2007, 01:56
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