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Odd/Even [#permalink]
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03 Oct 2008, 10:54
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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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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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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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^(n1) +1)/2 then you get 2^(n1) + 1 ..i.e even+odd=odd...always



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Re: Odd/Even [#permalink]
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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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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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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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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^(n1) +1)/2 then you get 2^(n1) + 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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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=4 124=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^(n1) +1)/2 then you get 2^(n1) + 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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05 Oct 2008, 12:23
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My approach is similar witha different flavor.
If ab 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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21 Oct 2008, 09:42
missed this one earlier. 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 = 244 = 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=4 124=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^(n1) +1)/2 then you get 2^(n1) + 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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21 Oct 2008, 16:18
scthakur wrote: My approach is similar witha different flavor.
If ab 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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22 Oct 2008, 01:04
LiveStronger wrote: scthakur wrote: My approach is similar witha different flavor.
If ab 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










