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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 co [#permalink]

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18 Nov 2009, 21:37
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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
[Reveal] Spoiler: OA
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Re: x is the sum of y consecutive integers. w is the sum of z co [#permalink]

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18 Nov 2009, 23:37
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

Case 1: x = w
if y = [2,3] z = [5]
5=5, TRUE

Case 2: x > w
if y = [2,3] z = [1]
5>1, TRUE

Case 3: x/y is an integer
if y = [2,3,4] x = 9
9/3 is an integer, TRUE

Case 4: w/z is an integer
if z = [2] w = 2
2/1 is an integer, TRUE

Case 5: x/z is an integer
if x = 9, z = [2]
9/1 is an integer, TRUE

Am I doing something wrong here?
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Re: x is the sum of y consecutive integers. w is the sum of z co [#permalink]

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19 Nov 2009, 04:07
h2polo 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

Case 1: x = w
if y = [2,3] z = [5]
5=5, TRUE

Case 2: x > w
if y = [2,3] z = [1]
5>1, TRUE

Case 3: x/y is an integer
if y = [2,3,4] x = 9
9/3 is an integer, TRUE

Case 4: w/z is an integer
if z = [2] w = 2
2/1 is an integer, TRUE

Case 5: x/z is an integer
if x = 9, z = [2]
9/1 is an integer, TRUE

Am I doing something wrong here?

well well well...I came to 3 taking y as a multiple of 2 (considering y=2z and y,z both are integers)
yet I may be wrong...

Last edited by kp1811 on 19 Nov 2009, 05:28, edited 1 time in total.
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Re: x is the sum of y consecutive integers. w is the sum of z co [#permalink]

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19 Nov 2009, 04:53
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For a set of 'n' consecutive integers, the sum of integers is :
(a) always a multiple of 'n' when n is ODD
(b) never a multiple of 'n' when n is EVEN
(Note : Try it out!)

From y = 2z, we can conclude that x is the sum of consecutive integers for an EVEN number of terms since y will always be even.

Thus x/y can never be an integer.

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16 Dec 2009, 03:08
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delta09 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

C

a) x= w if x = (1+2+3+4+5+6) = 21 and w = (6+ 7+8)=21.
b) x>w if x= (1+2+3+4+5+6) = 21 and w = (4+5+6)= 15
c) x/y => x = (1+2+3+4+5+6) = 21 and y = 6 then x/y is NOT an integer
=> x = (1+2+3+4) and y = 4 then x/y is NOT an integer
d) w/z =>w = (6+ 7+8)=21 and z=3 then w/z = 3
e) x/z => x = (1+2+3+4+5+6) = 21 and z = 3 then x/z = 7
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16 Dec 2009, 03:11
delta09 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

IMO (C).

y will always be even since y=2z

assume 6 numbers
n-2 n-1 n n+1 n+2 n+3
x=6n+3, y=6, z=3

x/z = 2n+1

x/y = n+1/2 ==> cannot be integer
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02 Jan 2010, 14:08
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OA is C

sum of even no is not always odd.

consider 1+2+3+4=10 which is even

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.

consider above example 10 is not multiple of 4.

But 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.

1+2+3=6 multiple of 3
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31 Jan 2010, 13:02
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delta09 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

since y = 2z... it is even...

x is the sum of y (even no) of integers....

Hence x/y would never be an integer....

Therefore C....

Note
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.
But 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.

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24 Dec 2010, 07:49
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nonameee 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

Quote:
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.
But 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.

• If $$k$$ is odd, the sum of $$k$$ consecutive integers is always divisible by $$k$$. Given $$\{9,10,11\}$$, we have $$k=3$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If $$k$$ is even, the sum of $$k$$ consecutive integers is never divisible by $$k$$. Given $$\{9,10,11,12\}$$, we have $$k=4$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of $$k$$ consecutive integers is always divisible by $$k!$$, so by $$k$$ too. Given $$k=4$$ consecutive integers: $$\{3,4,5,6\}$$. The product of 3*4*5*6 is 360, which is divisible by 4!=24.
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11 May 2012, 03:36
The question can be restated as:
If x is the sum of y consecutive integers, and w is the sum of y/2 consecutive integers, and neither of y or z is 0, then which of the following can NEVER be true?

Note that since y>0 and z=y/2 and z>0 and both y and z are integers, y will always be an even integer.

(a) x=w. This is certainly possible. Take x=-2,-1,0,1,2,3 and w=0,1,2
(b) x>w. This is certainly possible. Take x=-2,-1,0,1,2,3 and w=-2,-1,0
(c) Correct answer. You can never make a case that satisfies this. Why?
Sum of the AP = (y/2) [2a+(y-1)*1]
If y = 2k as y is always even, Sum = x = k [2a+ (2k-1)]
If we divide this by y, i.e. 2k, we get x/y = k[2a+2k-1]/2k. This is an odd number divided by an even number, and so can never be an integer.
(d) This is certainly possible. Evaluates to [2a+(z-1)]/2, which can be true if z is odd.
(e) This is certainly possible. Take x=-2,-1,0,1 and then z=2, and x/z=-1, which is an integer.
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Re: from manhattan cat 2 ps 28 [#permalink]

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24 Jun 2012, 19:55
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.

This is Mahattan's explanation.

Is there any other ways to solve this problem easily?
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Re: from manhattan cat 2 ps 28 [#permalink]

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24 Jun 2012, 22:03
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eybrj2 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

This is how you can work it out:

x = sum of y con. integers
w = sum of z con. integers

Plug in values such that most of these are easily made true.
Since z is in the denominator, I will try to put z = 1 to get w/z and x/z as integers without any complications.

If z = 1, y = 2
w = any one integer
x = sum of 2 consecutive integers
w/z and x/z will be integers so d and e can be true.

x = w is obviously possible. Say w = 9 and x = 4+5
x > w is also possible say if w = 8 while x = 4+5
Hence both a and b can be true.

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28 Aug 2012, 06:57
hrish88 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

I will post OA later.

Basically C is the average of even number of consecutive terms ,

average of even number of consecutive terms is never an integer ( 1,2,3,4 or 4,5,6,7)

good to know info.
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Re: x is the sum of y consecutive integers. w is the sum of z [#permalink]

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28 Aug 2012, 09:09
Karishma..is not diffcult way to do it?? to plug in value.. that will take time.

what i have read in Mgmat Number properties..

some of even numbers will never be divisble by even num i think m saying rite ..

y is a even num ..y=2z.. that mean y is multiple of 2.. and x will be even..both even numbers.. so can not be an integer..
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Re: x is the sum of y consecutive integers. w is the sum of z [#permalink]

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17 Oct 2012, 15:08
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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

x = w
x > w
x/y is an integer
w/z is an integer
x/z is an integer

average1= x/y
average2=w/z

y=2z,

implies that for every value of z y is even.
ie.
z=even say 4
y=2*4 =6 (even)

z=odd say 3

y=2*3 =6 (even)

the average of even numbers is not integer , so the answer is C.
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Re: x is the sum of y consecutive integers. w is the sum of z [#permalink]

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25 Oct 2012, 04:19
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delta09 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

For these kind of questions, we need to parse through the options to find an option which can NEVER be true. There are three scenarios possible with any option:
1. We can easily find out that this is possible. Then, we can eliminate this option and move forward.
2. We can easily find out that the option is never true. Thus, we can stop here because this is the answer.
3. Nothing can be easily inferred. We need to plug in values and do some reasoning to figure out. These options can be skipped for the moment to come back later, if we don't find an answer among other options. We want to SAVE our time.

Here is my attempt at this:
A. x = w - just looking at it, i can't say if this is possible or not. To find so will take time.I skip it to come back later
B. x > w - Easily possible is x is the sum of first 2 integers and w is the sum of first 1 integer
C. x/y is an integer - come back later
D. w/z is an integer - easily possible if z=1, w=1
E. x/z is an integer[/quote] - easily possible if z=1, x=3

now, we are left with A and C options. I begin with option C.

x=sum of consecutive y (or 2z) integers = n + (n+1) + (n+2) +....+ (n+2z-1) = 2zn + (1+2+3+...+2z-1) = 2zn + 2z*(2z-1)/2 (using formula for sum of first n natural numbers)
=> x/y or x/2z = n + (2z-1)/2
we know n is an integer. The other component can never be an integer, since the numerator is always odd and denominator is 2. So, x/y cannot be an integer.

Cheers,
CJ
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Re: x is the sum of y consecutive integers. w is the sum of z [#permalink]

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15 Aug 2013, 06:25
delta09 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

The fact is that y is always even and x is the sum of this y consecutive numbers.

like, x = 1+2+3+4 here y=4
again x= 6+7+8+9+10+11 and y=6
or x=1+2 and y=2 everywhere x/y = fraction
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Re: x is the sum of y consecutive integers. w is the sum of z co [#permalink]

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28 Sep 2013, 22:34
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Re: x is the sum of y consecutive integers. w is the sum of z [#permalink]

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05 Mar 2014, 23:49
sum of Y consecutive values would be (2a1+(y-1)d)*y/2. since they are consecutive, d=1. so we have an equation (2a1+(y-1))*y/2
if we devide it by Y, we are left with (2a1+(y-1))/2=a1+(y-a)/2
where y-1 is and odd number and will not be devisible by 2, so it is not an integer
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07 Mar 2015, 21:57
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