Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

x is the sum of y consecutive integers. w is the sum of z co [#permalink]

Show Tags

18 Nov 2009, 21:37

3

This post received KUDOS

21

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

34% (02:18) correct
66% (01:27) wrong based on 615 sessions

HideShow timer Statistics

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

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

Show Tags

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

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

Show Tags

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.

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

Show Tags

19 Nov 2009, 04:53

20

This post received KUDOS

14

This post was BOOKMARKED

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.

Answer : C _________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

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

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 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. _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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

Can someone please prove?

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. _________________

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. _________________

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?

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.

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)

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

Show Tags

17 Oct 2012, 15:08

1

This post received KUDOS

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.

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

Show Tags

25 Oct 2012, 04:19

1

This post received KUDOS

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.

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

Show Tags

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 _________________

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

Show Tags

28 Sep 2013, 22:34

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

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

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

Show Tags

07 Mar 2015, 21:57

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...