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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] New post 25 Nov 2008, 05:00
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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

I dont understand the answer choices here.
If I substitute 8 and 4 for a and b, i get both D&E as odd.
If I assume D&E to be even, both A and B need to be odd.
I just gave up and guessed C (not saying C is the OA)
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 07:25
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twilight 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

I dont understand the answer choices here.
If I substitute 8 and 4 for a and b, i get both D&E as odd.
If I assume D&E to be even, both A and B need to be odd.
I just gave up and guessed C (not saying C is the OA)


7-t66732
I think this prob and yours are the same.
a/b is even so a must be even
a+b is even so b must be even
a/b is even while b is even so a must be a multiple of 4.
a is a multiple of 4 => a+2 is not a multiple of 4. => (a+2)/2 must be odd.
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 05:33
twilight 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

I dont understand the answer choices here.
If I substitute 8 and 4 for a and b, i get both D&E as odd.
If I assume D&E to be even, both A and B need to be odd.
I just gave up and guessed C (not saying C is the OA)


I do not have the answer yet but I would like to comment on your reasoning. I think D and E can be odd or even, even when a and b are even. If you take a = 6, then you will have (6+2)/2 = 4 even. Or you can take b = 10, then you will have (10+2)/2 = 6 even.
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 05:36
Agree with you.
What I was poitning out there (with the 8 and 4 example) was that for two even numbers that satisfy the question, I get a situation where two of the choices are potentials answers.

---edit---
but I see the flaw in my reasoning here, since the question asks for a 'must be odd' response, so substitution is not the right way to go.
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 05:48
good question, and it initially confussed me :roll:

to satisfy a-b and a/b are both even integers, a and b should be both even.

A - a can be 2 (a/2=1) or 4 (a/2=2) Out
B - the same as A Out
C - (4+2)/2 = 3 or (8+4)/2=6
E - The same as in A,B

Thus, only D satisfies, because a can't be equal to 2

Last edited by atletikos on 25 Nov 2008, 12:27, edited 1 time in total.
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 06:29
Interesting answer, thanks for that.
I did not consider that 'a' must not be 2, which is true and your explantion makes sense.

However, theres nothing stopping a from being say a 6 - and in that case (a+2)/2 would be even, and hence would not agree with the requirement of 'must be odd'..

what say?
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 07:18
I get D too...basically A >2 and could be of the form 2*N...so 2*n/2=n+1 is always odd.
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 07:38
fresinha12 wrote:
I get D too...basically A >2 and could be of the form 2*N...so 2*n/2=n+1 is always odd.


But if N is odd (which it can be), then 2*n/2=n+1 will be even?
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 07:40
DavidArchuleta wrote:

7-t66732
I think this prob and yours are the same.
a/b is even so a must be even
a+b is even so b must be even
a/b is even while b is even so a must be a multiple of 4.
a is a multiple of 4 => a+2 is not a multiple of 4. => (a+2)/2 must be odd.


ah yes!
thats perfect, thanks.
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 07:41
twilight wrote:
fresinha12 wrote:
I get D too...basically A >2 and could be of the form 2*N...so 2*n/2=n+1 is always odd.


But if N is odd (which it can be), then 2*n/2=n+1 will be even?



this is not possible sinc a/b is an even integer which means N is even..
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Re: Even/odd integers [#permalink] New post 25 Nov 2008, 20:31
DavidArchuleta wrote:
twilight 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

I dont understand the answer choices here.
If I substitute 8 and 4 for a and b, i get both D&E as odd.
If I assume D&E to be even, both A and B need to be odd.
I just gave up and guessed C (not saying C is the OA)


7-t66732
I think this prob and yours are the same.
a/b is even so a must be even
a+b is even so b must be even
a/b is even while b is even so a must be a multiple of 4.
a is a multiple of 4 => a+2 is not a multiple of 4. => (a+2)/2 must be odd.



+1 David. Good explanation
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Re: Even/odd integers [#permalink] New post 28 Nov 2008, 19:08
You may use substitution:

Initially substitute a and b with 8 and 2 respectively.
a = 8
b = 2

8 - 2 = even number
8/2 = even number

so these numbers satisfy requirements

a) a/2: 8/2 = 4 EVEN

b) b/2: 4/2 = 2 EVEN

c) (a+b)/2: (8+4)/2 = 6 EVEN

d) (a+2)/2: (8+2)/2 = 5 ODD
substitute another set (a=4 & b=2) for check
(4+2)/2 = 3 Again, ODD

e) (b+2)/2: (4+2)/2 = 3 ODD
substitute another set (a=4 & b=2) for check
(2+2)/2 = 2 EVEN

ANSWER: D
Re: Even/odd integers   [#permalink] 28 Nov 2008, 19:08
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