|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 10 Mar 2008
Posts: 377
Followers: 3
Kudos [?]:
19
[0], given: 0
|
I cant figure out which one of these will be ODD. Every single option turns out to be even.
Attachments
question.jpg [ 11.43 KiB | Viewed 558 times ]
|
|
|
|
|
|
|
Manager
Joined: 22 Sep 2008
Posts: 124
Followers: 1
Kudos [?]:
34
[0], given: 0
|
c ...
a+b/2 will be right answer
|
|
|
|
|
|
Current Student
Joined: 28 Dec 2004
Posts: 3437
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
135
[0], given: 2
|
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
|
|
|
|
|
|
Intern
Joined: 29 Sep 2008
Posts: 47
Followers: 0
Kudos [?]:
8
[0], given: 0
|
a combination of calculation and intuition and substitution suggests to me it should be D.
Please confirm.
|
|
|
|
|
|
Director
Joined: 27 Jun 2008
Posts: 551
WE 1: Investment Banking - 6yrs
Followers: 1
Kudos [?]:
39
[0], given: 92
|
I get D
a = 12
b = 6
I get b,c & d odd
a = 8
b = 4
d is odd
|
|
|
|
|
|
Intern
Joined: 29 Sep 2008
Posts: 47
Followers: 0
Kudos [?]:
8
[0], given: 0
|
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.
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2528
Followers: 41
Kudos [?]:
365
[0], given: 19
|
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.
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Current Student
Joined: 28 Dec 2004
Posts: 3437
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
135
[0], given: 2
|
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. |
|
|
|
|
|
SVP
Joined: 17 Jun 2008
Posts: 1592
Followers: 7
Kudos [?]:
132
[1] , given: 0
|
1
This post received
KUDOS
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.
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2528
Followers: 41
Kudos [?]:
365
[0], given: 19
|
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 = 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. _________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Senior Manager
Joined: 21 Apr 2008
Posts: 275
Location: Motortown
Followers: 0
Kudos [?]:
65
[0], given: 0
|
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.
|
|
|
|
|
|
SVP
Joined: 17 Jun 2008
Posts: 1592
Followers: 7
Kudos [?]:
132
[0], given: 0
|
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
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
