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x is the sum of y consecutive integers. w is the sum of z

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x is the sum of y consecutive integers. w is the sum of z [#permalink] New post 17 Jan 2008, 11:08
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x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

A.x = w
B.x > w
C.x/y is an integer
D.w/z is an integer
E.x/z is an integer
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Re: PS - Consecutive integers [#permalink] New post 17 Jan 2008, 11:33
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tough to be solved for 5min. :wall
I guess A.
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Re: PS - Consecutive integers [#permalink] New post 17 Jan 2008, 13:08
JCLEONES wrote:
x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

A.x = w
B.x > w
C.x/y is an integer
D.w/z is an integer
E.x/z is an integer


in 2 min elim BCD. B is obvs that x can be greater than w.
C: just make a scenario 1,2,3,4,5 15/5 OK
D: just look at C.

Ok now down to A and E. Here I just guessed b/c I ran outta time. I said A

Now lets relook at this x/z ok lets say we have 2 as z and 4 as y 1+2+3+4. =10 =x 10/2 is an integer
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Re: PS - Consecutive integers [#permalink] New post 17 Jan 2008, 14:40
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The answer is C.

x = k + (k+1) + ... + (k+2*z-1) = (2*k+2*z-1)*2z/2 = (2*k+2*z-1)*z

y = 2*z

x/y = (k+z)-1/2 , can never be an integer
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Re: PS - Consecutive integers [#permalink] New post 17 Jan 2008, 17:20
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A. x = w

x = 1+2 = 3
w = 3 = 3

B. x = 5+6+7+8 = 26
w = 5+6 = 11

C. x = -1+0+1 = 0
y = 3

0/3 = 0 = integer

OR

x = 0+1+2 = 3
y = 3
3/3 = 1 = integer

D. same thing as C

E. x = 2+3+4+5 = 14
z = 2
14/2 = 7


EDIT: I found the mistake. If y = 2z then y must be an even number. My examples all use 3 numbers for y, making them impossible. Under these constraints C is the answer.

Answer C
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Re: PS - Consecutive integers [#permalink] New post 18 Jan 2008, 07:34
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Yes,you all deserve +1!
The OA is indeed C.
Please read the following as it maybe helpful for all:

´´For any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms. For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3. For any set of consecutive integers with an even number of terms, the sum of the integers is never a multiple of the number of terms. For example, the sum of 1, 2, 3, and 4 (four consecutives -- an even number) is 10, which is not a multiple of 4.

The question tells us that y = 2z, which allows us to deduce that y is even. Since y is even, then the sum of y integers, x, cannot be a multiple of y. Therefore, x/y cannot be an integer; choice C is the correct answer. ´´



walker wrote:
Yes, you guys are right.
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Re: PS - Consecutive integers [#permalink] New post 18 Jan 2008, 08:07
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JCLEONES wrote:
´´For any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms. For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3. For any set of consecutive integers with an even number of terms, the sum of the integers is never a multiple of the number of terms. For example, the sum of 1, 2, 3, and 4 (four consecutives -- an even number) is 10, which is not a multiple of 4.


Thanks. Very helpful. I remember that I saw other problem that test this feature. It was difficult for me to solve it in 2 min....
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Re: PS - Consecutive integers   [#permalink] 18 Jan 2008, 08:07
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