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

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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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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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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

Wrong answer. Try again

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

It should be E then (by trial and error)..
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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

Hence D is the answer
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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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Re: positive integers - PS [#permalink]

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06 Aug 2008, 16:53
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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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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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Re: positive integers - PS [#permalink]

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07 Aug 2008, 16:58
Oh thanks
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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.
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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

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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,

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

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