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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

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.

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.

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.

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.

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