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# PS: Except

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SVP
Joined: 29 Aug 2007
Posts: 2419

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11 Dec 2008, 15:28
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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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Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown

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11 Dec 2008, 17:34
GMAT TIGER 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

y is an even number, and sum of even consecutive numbers can't be divided by that even number, so I would go with C
Intern
Joined: 22 Dec 2007
Posts: 5

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12 Dec 2008, 02:23
Why can't the sum of even consecutive integers be divisible by even number???

2+3+4+5 = 14

is 14 not devisible by even number, such as 2 or 4?
Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown

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12 Dec 2008, 04:48
I am sorry for the confusion, I wrote, "and sum of even consecutive numbers can't be divided by that even number", and I realised it is not clear.

what I meant was:

Lets take your example: 2+3+4+5 = 14

14 here is a sum of 4 consecutive inetegers, and 14 can't be divided by 4

If we substitute w = 14, y = 4 w/y can't be an integer.

What is the OA Tiger ?
Intern
Joined: 26 Sep 2008
Posts: 19

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12 Dec 2008, 20:09
I agree with C.
we have
x = y.k + y(y-1)/2 in which k is the first element of y consecutive integers.
w= z.l + z(z-1)/2 in which k is the first element of z consecutive integers.
y=2z so
x=2z.k + 2z(2z-1)/2
and x/y = k + (2z-1)/2 cannot be an integer.
Senior Manager
Joined: 30 Nov 2008
Posts: 478
Schools: Fuqua

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Updated on: 09 Mar 2009, 05:51
I would also go with C. But my explanation is slightly different.

Given , Sum of y consecutive integers is X. Sum of z consecutive integers is W.

Also y = 2z which means we can conclude that y is always even and w can be even or odd.

When I scanned thru the options, I see that options C and D have some relation specified which is nothing but the avg.

If I look at option C which is X/y, it is nothing but the Avg of y consecutive Integers. For even number is consecutive integers, average cannot be an integer since it always is the avg of the 2 middle integers. Hence we can conclude that X / y can never be an integer.

If I loot at option D, which is W/z, again avg of z consecutive integers, can be an integer if z is odd or cannot be an integer if z is even. SO there is a possibility of W/z to be an integer.

Originally posted by mrsmarthi on 13 Dec 2008, 12:49.
Last edited by mrsmarthi on 09 Mar 2009, 05:51, edited 1 time in total.
Intern
Joined: 19 Jun 2008
Posts: 20

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13 Dec 2008, 14:08
ngotuan wrote:
I agree with C.
we have
x = y.k + y(y-1)/2 in which k is the first element of y consecutive integers.
w= z.l + z(z-1)/2 in which k is the first element of z consecutive integers.
y=2z so
x=2z.k + 2z(2z-1)/2
and x/y = k + (2z-1)/2 cannot be an integer.

Would you mind explaining the summation formulas that you used?
SVP
Joined: 07 Nov 2007
Posts: 1728
Location: New York

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13 Dec 2008, 19:55
GMAT TIGER 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

x= sum of y consecutive integers.
= (k+1)+(k+2)...+(k+y) (here K can be any integer (+ve or -ve or zero)
= yk+ (1+2+3.. y)
= yk+ y(y+1)/2

x/y = k+(y+1)/2 = k+(2z+1)/2 --> can't be integer because (2z+1) odd number divided by 2 is always fraction number.
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Director
Joined: 04 Jan 2008
Posts: 818

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09 Mar 2009, 03:45
Awesome and thanks

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Re: PS: Except &nbs [#permalink] 09 Mar 2009, 03:45
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