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 350,000 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:

If a and b are positive integers such that a b and a/b are [#permalink]
06 Aug 2008, 11:50

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

50% (01:48) correct
50% (00:00) wrong based on 3 sessions

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

Re: positive integers - PS [#permalink]
06 Aug 2008, 11:58

chan4312 wrote:

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

Re: positive integers - PS [#permalink]
06 Aug 2008, 12:55

chan4312 wrote:

x2suresh wrote:

chan4312 wrote:

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

Re: positive integers - PS [#permalink]
06 Aug 2008, 13:31

chan4312 wrote:

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

IMO D

Take a = 4,5,6 and b = 2,1,2; perform the operations accordingly (a-b) = 2, 4, 4, all even (a/b) = 2,5,3, even and odd Now only even A and B gives even for both the operations; The numbers that satisfy are a = 4,8 b = 2,4

Re: positive integers - PS [#permalink]
06 Aug 2008, 13:40

a-b = even . i.e either both are EVEN or ODD ------- 1

a/b=even; this suggests 'a' has to be even and 'b' might be odd or even ----- 2

from 1 and 2 both a and b are even

Now at the options :

1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO

Re: positive integers - PS [#permalink]
06 Aug 2008, 15:53

3

This post received KUDOS

chan4312 wrote:

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

a/b is even indicates a is even ,a-b is even integer only when a and b are both even integers.

also most importantly a must be multiple of 4 so that a/b is even inspite of b being even

hence say a=4k where k is integer.

now eliminate (a) a/2 is even eliminate (b) b/2 is even or odd say b=4 ,say b=6 then b/2 varies eliminate (c)a+b /2 is again even or odd (D)(a+2)/2 is always odd -> (a+2)/2 = (4k+2)/2 =2k+1 => hence always odd eliminate (E)b+2 /2 is even or odd

Re: positive integers - PS [#permalink]
30 Aug 2011, 21:01

stallone wrote:

a-b = even . i.e either both are EVEN or ODD ------- 1

a/b=even; this suggests 'a' has to be even and 'b' might be odd or even ----- 2

from 1 and 2 both a and b are even

Now at the options :

1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO

What if we take a=6 then option 4 would become----> 6+2/2 which is even...

Re: positive integers - PS [#permalink]
30 Aug 2011, 22:38

DeeptiM wrote:

stallone wrote:

a-b = even . i.e either both are EVEN or ODD ------- 1

a/b=even; this suggests 'a' has to be even and 'b' might be odd or even ----- 2

from 1 and 2 both a and b are even

Now at the options :

1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO

What if we take a=6 then option 4 would become----> 6+2/2 which is even...

pls help me understand this.

Hello,

You cannot consider a= 6 because you cannot find another integer "b" such that a/b is even and a - b is also even

The possible values for a,b = {(4,2),(8,2),(8,4),(4,4)} and so on

Re: positive integers - PS [#permalink]
30 Aug 2011, 22:50

DeeptiM wrote:

stallone wrote:

a-b = even . i.e either both are EVEN or ODD ------- 1

a/b=even; this suggests 'a' has to be even and 'b' might be odd or even ----- 2

from 1 and 2 both a and b are even

Now at the options :

1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO

What if we take a=6 then option 4 would become----> 6+2/2 which is even...

pls help me understand this.

Hello,

Just to add

1. a and b need to be even

2. the number of 2's in the prime factorization of "a" must be greater than number of 2's in the prime factorization of "b"

regards Raghav.V Consider kudos if you find my post helpful:

Here, a-b = +ve integer and a/b = +ve integer In order to satisfy both condition, a and b both must be even integer. Given the answer choice, it is clear that the average of two even integer is always Odd.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...