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Odd/Even

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Odd/Even [#permalink]

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New post 03 Oct 2008, 10:54
I cant figure out which one of these will be ODD. Every single option turns out to be even.

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Re: Odd/Even [#permalink]

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New post 03 Oct 2008, 10:58
c ...

a+b/2 will be right answer
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Re: Odd/Even [#permalink]

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New post 03 Oct 2008, 11:03
i get d..

a+2/2 will be odd..

from the stem we know both a and b are even..

we also know that a=b*even which means a will always b divisible by 4..

so lets pick 8 +2=10/2=5 you cant pick something like 12 since it violates the stem..

basically a=2^n...

2^n +2=2(2^(n-1) +1)/2 then you get 2^(n-1) + 1 ..i.e even+odd=odd...always
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Re: Odd/Even [#permalink]

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New post 03 Oct 2008, 11:03
a combination of calculation and intuition and substitution suggests to me it should be D.
Please confirm.
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Re: Odd/Even [#permalink]

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New post 03 Oct 2008, 11:08
I get D

a = 12
b = 6

I get b,c & d odd

a = 8
b = 4

d is odd
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Re: Odd/Even [#permalink]

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New post 03 Oct 2008, 11:15
aim2010 wrote:
a combination of calculation and intuition and substitution suggests to me it should be D.
Please confirm.

looking at fresinha post i think i have a mathematical proof for D.

it can be inferred that a can be expressed as 4x and b as 2y. just to drive home the "must" point in the question, it is possible that b can also take a form of 4y, but we are taking the worst case scenario, hence a will ALWAYS have a factor of 4, not b.

(a+2)/2=(4x+2)/2=2x+1 --> always odd none of the other options satisfy with such certainty.
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Re: Odd/Even [#permalink]

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New post 03 Oct 2008, 11:28
fresinha12 wrote:
i get d..

a+2/2 will be odd..

from the stem we know both a and b are even..

we also know that a=b*even which means a will always b divisible by 4..

so lets pick 8 +2=10/2=5 you cant pick something like 12 since it violates the stem..

basically a=2^n...

2^n +2=2(2^(n-1) +1)/2 then you get 2^(n-1) + 1 ..i.e even+odd=odd...always


Agree with D but the statement that "basically a=2^n..." is not true cuz "a" has to be a multiple of 4. in that case "a" could be 4 or 8 or 12. If a is 12, then it doesnot fit anywhere on the expression that a = 2^n.... so a = 2^n is not correct.

the strongest reason why (a+2)/2 is odd is that a must be a multiple of 4.
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Re: Odd/Even [#permalink]

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New post 05 Oct 2008, 10:51
I dont think you can have any odd prime factors for a..

lets say for argument sake..a=12 and b=4 12-4=8 but 12/4=3 so you see you can have a multiple of 4, but it wont meet the requirement laid out in the stem..therefore i feel..a at the least has to be 2^n ...

GMAT TIGER wrote:
fresinha12 wrote:
i get d..

a+2/2 will be odd..

from the stem we know both a and b are even..

we also know that a=b*even which means a will always b divisible by 4..

so lets pick 8 +2=10/2=5 you cant pick something like 12 since it violates the stem..

basically a=2^n...

2^n +2=2(2^(n-1) +1)/2 then you get 2^(n-1) + 1 ..i.e even+odd=odd...always


Agree with D but the statement that "basically a=2^n..." is not true cuz "a" has to be a multiple of 4. in that case "a" could be 4 or 8 or 12. If a is 12, then it doesnot fit anywhere on the expression that a = 2^n.... so a = 2^n is not correct.

the strongest reason why (a+2)/2 is odd is that a must be a multiple of 4.
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Re: Odd/Even [#permalink]

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New post 05 Oct 2008, 12:23
1
My approach is similar witha different flavor.

If a-b is even then both a and b are either odd or both are even.
If a/b is also even then both a and b cannot be odd.

Hence, combining both, a and b are even in such a way that a = (2x)b for x = 0,1,2,3,4......

Hence, a/2 will be xb and since b is even, xb will always be even and hence a/2 will also always be even and a/2 + 1 will always be odd.
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Re: Odd/Even [#permalink]

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New post 21 Oct 2008, 09:42
missed this one earlier. :oops:

Yes you can. Lets take a = 24, which is a multiple of 4 and has 3 as odd prime factor, and b = 4. then
a - b = 24-4 = 20
a/b = 24/4 = 6
(a+2)/2 = (24+2)/2 = 13

So it is not necessary for "a" to have its value a power of 2.




fresinha12 wrote:
I dont think you can have any odd prime factors for a..

lets say for argument sake..a=12 and b=4 12-4=8 but 12/4=3 so you see you can have a multiple of 4, but it wont meet the requirement laid out in the stem..therefore i feel..a at the least has to be 2^n ...

GMAT TIGER wrote:
fresinha12 wrote:
i get d..

a+2/2 will be odd..

from the stem we know both a and b are even..

we also know that a=b*even which means a will always b divisible by 4..

so lets pick 8 +2=10/2=5 you cant pick something like 12 since it violates the stem..

basically a=2^n...

2^n +2=2(2^(n-1) +1)/2 then you get 2^(n-1) + 1 ..i.e even+odd=odd...always


Agree with D but the statement that "basically a=2^n..." is not true cuz "a" has to be a multiple of 4. in that case "a" could be 4 or 8 or 12. If a is 12, then it doesnot fit anywhere on the expression that a = 2^n.... so a = 2^n is not correct.

the strongest reason why (a+2)/2 is odd is that a must be a multiple of 4.

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Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


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Re: Odd/Even [#permalink]

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New post 21 Oct 2008, 16:18
scthakur wrote:
My approach is similar witha different flavor.

If a-b is even then both a and b are either odd or both are even.
If a/b is also even then both a and b cannot be odd.

Hence, combining both, a and b are even in such a way that a = (2x)b for x = 0,1,2,3,4......

Hence, a/2 will be xb and since b is even, xb will always be even and hence a/2 will also always be even and a/2 + 1 will always be odd.


I will go with this approach, except that x can't be ZERO, since a and b are positive integers.
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Re: Odd/Even [#permalink]

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New post 22 Oct 2008, 01:04
LiveStronger wrote:
scthakur wrote:
My approach is similar witha different flavor.

If a-b is even then both a and b are either odd or both are even.
If a/b is also even then both a and b cannot be odd.

Hence, combining both, a and b are even in such a way that a = (2x)b for x = 0,1,2,3,4......

Hence, a/2 will be xb and since b is even, xb will always be even and hence a/2 will also always be even and a/2 + 1 will always be odd.


I will go with this approach, except that x can't be ZERO, since a and b are positive integers.


Thanks livestronger for pointing this out. I must improve upon reading the questions in full :oops:

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Re: Odd/Even   [#permalink] 22 Oct 2008, 01:04
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