Find all School-related info fast with the new School-Specific MBA Forum

It is currently 23 Oct 2014, 11:12

Going on right now!

Interview Invites from Chicago - Booth | Join the chat room for Live Updates.


Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

a is the sum of x consecutive positive integers. b is the

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Director
Director
avatar
Joined: 13 Nov 2003
Posts: 801
Location: BULGARIA
Followers: 1

Kudos [?]: 21 [0], given: 0

a is the sum of x consecutive positive integers. b is the [#permalink] New post 15 May 2006, 23:49
a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

(A) x = 2; y = 6
(B) x = 3; y = 6
(C) x = 7; y = 9
(D) x = 10; y = 4
(E) x = 10; y = 7

No OA
Senior Manager
Senior Manager
avatar
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 16 May 2006, 03:50
BG wrote:
a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

(A) x = 2; y = 6
(B) x = 3; y = 6
(C) x = 7; y = 9
(D) x = 10; y = 4
(E) x = 10; y = 7

No OA


D it is

Sum = average * number

the average of even number of consecutive integers is always decimal
while the average of odd number of con. int. is always integer

assume, the average of X a(x) and the average of Y a(y)

A) a(x)*2 = a(y)*6 => a(x) (decimal) = 3*a(y)(decimal) OK
B) a(x)*3 = a(y)*6 => a(x) (integer) = 2*a(y)(2*decimal = integer) OK
C) a(x)*7 (integer)= a(y)*9(integer) OK
D) a(x)*10 = a(y)*4 => 5*a(x)(decimal) = 2*a(y) ( 2 * decimal = integer ) DECIMAL != INTEGER
E) a(x)*10 ( 10 * decimal = integer ) = a(y)*7 (integer) OK
Director
Director
avatar
Joined: 24 Oct 2005
Posts: 662
Location: London
Followers: 1

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 16 May 2006, 04:09
deowl wrote:
BG wrote:
a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

(A) x = 2; y = 6
(B) x = 3; y = 6
(C) x = 7; y = 9
(D) x = 10; y = 4
(E) x = 10; y = 7

No OA


D it is

Sum = average * number

the average of even number of consecutive integers is always decimal
while the average of odd number of con. int. is always integer

assume, the average of X a(x) and the average of Y a(y)

A) a(x)*2 = a(y)*6 => a(x) (decimal) = 3*a(y)(decimal) OK
B) a(x)*3 = a(y)*6 => a(x) (integer) = 2*a(y)(2*decimal = integer) OK
C) a(x)*7 (integer)= a(y)*9(integer) OK
D) a(x)*10 = a(y)*4 => 5*a(x)(decimal) = 2*a(y) ( 2 * decimal = integer ) DECIMAL != INTEGER
E) a(x)*10 ( 10 * decimal = integer ) = a(y)*7 (integer) OK

Hey, that is excellent !!
I was lost trying to figure out how to do this. :good
Manager
Manager
avatar
Joined: 07 Sep 2004
Posts: 61
Followers: 1

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 May 2006, 04:10
For part a)
a=10+11=b=1+2+3+4+5+6=21
for part b)
a= 6+7+8=1+2+3+4+5+6=21
for part c)
a=6+7+8+9+10+11=3+4+5+6+7+8+9+10+11=63
for part d) impossible
for part e)
a=6+7+8+9+10+11+12+13+14+15=12+13+14+15+16+17+18=105
answer is d)

further explanation in support for d
a= n+(n+1)+(n+2)+(n+3)=4n+6
b=m+ (m+1)+ (m+2)+……(m+9)= 10m+45

4n+6=10m+45
4n=10m+39
4n is even but 10m+39 is odd , so it is impossible
Intern
Intern
avatar
Joined: 10 Jan 2006
Posts: 25
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 May 2006, 08:00
deowl wrote:
BG wrote:
a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

(A) x = 2; y = 6
(B) x = 3; y = 6
(C) x = 7; y = 9
(D) x = 10; y = 4
(E) x = 10; y = 7

No OA


D it is

Sum = average * number

the average of even number of consecutive integers is always decimal
while the average of odd number of con. int. is always integer

assume, the average of X a(x) and the average of Y a(y)

A) a(x)*2 = a(y)*6 => a(x) (decimal) = 3*a(y)(decimal) OK
B) a(x)*3 = a(y)*6 => a(x) (integer) = 2*a(y)(2*decimal = integer) OK
C) a(x)*7 (integer)= a(y)*9(integer) OK
D) a(x)*10 = a(y)*4 => 5*a(x)(decimal) = 2*a(y) ( 2 * decimal = integer ) DECIMAL != INTEGER
E) a(x)*10 ( 10 * decimal = integer ) = a(y)*7 (integer) OK


Although (D) is correct the assertion below is not.
5*a(x)(decimal) = 2*a(y) Integer => DECIMAL = Integer

if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18

I solved it a slightly different way.

Sum of X cons integers = Kx + x(x+1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y+1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*11/2 = 10K + 55
SUM2 = 4K1 + 4*5/2 = 4K1 + 10

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.
Intern
Intern
avatar
Joined: 12 Nov 2003
Posts: 19
Followers: 0

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 16 May 2006, 08:13
saha wrote:
if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18



the sum of x consecutive positive integers. b is the sum of y consecutive positive integers

i think your example is not right.
Intern
Intern
avatar
Joined: 04 May 2006
Posts: 40
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 May 2006, 08:14
Saha, you are close. But i believe the formula would be

Sum of X cons integers = Kx + x(x-1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y-1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*9/2 = 10K + 45
SUM2 = 4M + 4*3/2 = 4M + 6

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.
Intern
Intern
avatar
Joined: 10 Jan 2006
Posts: 25
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 May 2006, 09:37
kpoxa wrote:
saha wrote:
if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18



the sum of x consecutive positive integers. b is the sum of y consecutive positive integers

i think your example is not right.


a(x) is the average of X consecutive integers, a(y) is the averge of Y consecutive integers according to the example given above ? What's wrong ?
Intern
Intern
avatar
Joined: 10 Jan 2006
Posts: 25
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 May 2006, 09:39
mattflow wrote:
Saha, you are close. But i believe the formula would be

Sum of X cons integers = Kx + x(x-1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y-1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*9/2 = 10K + 45
SUM2 = 4M + 4*3/2 = 4M + 6

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.


Yes, thanks for the correction. My formula would be correct if K+1 and K1+1 were the lowest, not K and K1.
VP
VP
User avatar
Joined: 29 Dec 2005
Posts: 1351
Followers: 7

Kudos [?]: 28 [0], given: 0

Re: sum of integers MNHTN [#permalink] New post 16 May 2006, 10:03
BG wrote:
a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

(A) x = 2; y = 6
(B) x = 3; y = 6
(C) x = 7; y = 9
(D) x = 10; y = 4
(E) x = 10; y = 7


(a) take any 6 consecutive integers (+ve) for y = 10, 11, 12, 13, 14 and 15. their sum is 75. divide it by 2, we get 37.5. so take immidiate integers above and below 37.5. they are 37 and 38 and their sum is75.

(b) take any 6 consecutive integers for y = 10, 11, 12, 13, 14 and 15. their sum is 75. divide it by 3, we get 25. so put 25 in the middle. the 3 consecutive integers are 24, 25 and 26. their sum is 75.

(c) take any 9 consecutive integers for y = 3, 4, 5, 6, 7, 8, 9, 10, and 11. their sum is 63. divide it by 7, we get 9. so put 9 in the middle of 7 consecutive integers. they are 6, 7, 8, 9, 10, 11, and 12. their sum is 63.

(d) take any 10 consecutive integers for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. their sum is 55. similarly take any 4 consecutive integers and add them up, the sum is even number/integer.

here we have interesting pattren with 10 and 4 consecutive integers. any 10 such integers have sum ending at 5 where as sum of 4 consecutive integers end at even integer. so these two consecutive integers never be equal.

(e) take any 10 consecutive integers for x = 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15. their sum is 105. divide it by 7, we get 15. so put 15 in the middle of 7 consecutive integers. they are 12, 13, 14, 15, 16, 17, and 18. their sum is 105.

it is D, however it is lengthy.
Senior Manager
Senior Manager
avatar
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 16 May 2006, 10:45
saha wrote:


Although (D) is correct the assertion below is not.
5*a(x)(decimal) = 2*a(y) Integer => DECIMAL = Integer

if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18




The decimal average can never be 0.4 or anything different from
.5 in tenths. And it will never give an integer when multiplied by 5.
So my assertion is perfectly correct.
Intern
Intern
avatar
Joined: 10 Jan 2006
Posts: 25
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 May 2006, 13:40
deowl wrote:
saha wrote:


Although (D) is correct the assertion below is not.
5*a(x)(decimal) = 2*a(y) Integer => DECIMAL = Integer

if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18




The decimal average can never be 0.4 or anything different from
.5 in tenths. And it will never give an integer when multiplied by 5.
So my assertion is perfectly correct.


I realised that later but it wasn't mentioned in that form earlier. its a good method i think.
Intern
Intern
avatar
Joined: 15 May 2006
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 18 May 2006, 07:05
mattflow wrote:
Saha, you are close. But i believe the formula would be

Sum of X cons integers = Kx + x(x-1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y-1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*9/2 = 10K + 45
SUM2 = 4M + 4*3/2 = 4M + 6

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.


I did the same way, but i dont agree with the first solution of sum=average*number
CEO
CEO
User avatar
Joined: 20 Nov 2005
Posts: 2922
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 15

Kudos [?]: 82 [0], given: 0

 [#permalink] New post 18 May 2006, 08:31
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always ODD : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008


Last edited by ps_dahiya on 18 May 2006, 12:24, edited 1 time in total.
Manager
Manager
avatar
Joined: 10 May 2006
Posts: 186
Location: USA
Followers: 1

Kudos [?]: 3 [0], given: 0

 [#permalink] New post 18 May 2006, 11:53
ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always EVEN : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD


Great shortcut! However, i think you made a typo in (A) --> a should be odd.
Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 383
Followers: 1

Kudos [?]: 18 [0], given: 0

 [#permalink] New post 18 May 2006, 12:09
ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always EVEN : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD


Yep I solved the same way!
CEO
CEO
User avatar
Joined: 20 Nov 2005
Posts: 2922
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 15

Kudos [?]: 82 [0], given: 0

 [#permalink] New post 18 May 2006, 12:25
tl372 wrote:
ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always EVEN : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD


Great shortcut! However, i think you made a typo in (A) --> a should be odd.


Yes that was a typo. Thanks buddy. I edited that.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Manager
Manager
avatar
Joined: 29 Apr 2006
Posts: 85
Followers: 1

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 18 May 2006, 18:16
ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always ODD : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD


Shouldn't (E) be: a is always odd, b maybe odd or even?
  [#permalink] 18 May 2006, 18:16
    Similar topics Author Replies Last post
Similar
Topics:
10 Experts publish their posts in the topic If y is the sum of x consecutive positive integers, and x>3, goodyear2013 5 24 Mar 2014, 14:34
5 Experts publish their posts in the topic If a and b are consecutive positive integers, and ab = 30x Maxirosario2012 2 30 Nov 2013, 18:37
21 Experts publish their posts in the topic If a is the sum of x consecutive positive integers. b is the devinawilliam83 10 05 Mar 2012, 23:36
6 Experts publish their posts in the topic If a and b are consecutive positive integers, and ab = 30x , 4test1 12 09 Nov 2009, 21:04
2 If the sum of 4 consecutive positive integers is divisible b tejal777 9 28 Sep 2009, 18:40
Display posts from previous: Sort by

a is the sum of x consecutive positive integers. b is the

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.