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# If a and b are positive integers such that a b and a/b are

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Manager
Joined: 04 Jun 2008
Posts: 157
If a and b are positive integers such that a b and a/b are [#permalink]

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06 Aug 2008, 12:50
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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
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Joined: 07 Nov 2007
Posts: 1799
Location: New York
Re: positive integers - PS [#permalink]

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06 Aug 2008, 12:58
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

C

a-b= 2k a/b= 2l -- > a= 2bl
(a+b)/2 --> 2bl+b/2 --> b/2(2l+1)
(2l+1) is always odd.. odd * b/2 --> odd
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Last edited by x2suresh on 06 Aug 2008, 13:39, edited 2 times in total.
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Location: New York
Re: positive integers - PS [#permalink]

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06 Aug 2008, 13:55
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

C

a-b= 2k a/b= 2l -- > a= 2bl
(a+b)/2 --> 2bl+b/2 --> b/2(2l+1)
(2l+1) is always odd.. odd * b/2 --> odd

oops you are right.. odd * even -- > even.

It should be E then (by trial and error)..
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Posts: 324
Re: positive integers - PS [#permalink]

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06 Aug 2008, 14:31
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

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Posts: 296
Re: positive integers - PS [#permalink]

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06 Aug 2008, 14:40
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
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Joined: 17 Jun 2008
Posts: 1381
Re: positive integers - PS [#permalink]

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06 Aug 2008, 16:53
3
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

IMO D
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Manager
Joined: 04 Jun 2008
Posts: 157
Re: positive integers - PS [#permalink]

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07 Aug 2008, 06:01
OA is D

Thank you all..
pretty good explanation spriya.!
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Posts: 1381
Re: positive integers - PS [#permalink]

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07 Aug 2008, 16:58
Oh thanks
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Posts: 258
Re: positive integers - PS [#permalink]

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30 Aug 2011, 22:01
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...

pls help me understand this.
Intern
Joined: 11 May 2011
Posts: 22
Re: positive integers - PS [#permalink]

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30 Aug 2011, 23:38
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

regards
Raghav.V

Consider kudos if you find my post helpful:
Intern
Joined: 11 May 2011
Posts: 22
Re: positive integers - PS [#permalink]

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30 Aug 2011, 23:50
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,

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:
Manager
Joined: 09 Jun 2011
Posts: 103
Re: positive integers - PS [#permalink]

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01 Sep 2011, 19:40
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.

i.e. (a+b)/2 is always odd
Re: positive integers - PS   [#permalink] 01 Sep 2011, 19:40
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