If a and b are positive integers such that a-b and a/b : PS Archive
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If a and b are positive integers such that a-b and a/b

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If a and b are positive integers such that a-b and a/b [#permalink]

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04 Jul 2005, 02:34
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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

Thanks...
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cheebinn

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04 Jul 2005, 04:38
But I plugged in values and got.....maybe someone can give a proper explanation
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04 Jul 2005, 05:10
I got D.

a must be a multiple of 4. hence (a + 2)/2 = (4k + 2)/2 = 2k + 1 which is always odd.

HMTG.
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04 Jul 2005, 06:28
Pluggin in numbers doesn't seem to solve this.
For a/b to be even, both a and b has to be even or a even(such as 10) and b odd (such as 5).

but 10 -5 is not even.
Is there a trick to this question?
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05 Jul 2005, 09:55
it looks like C, D, and E all can be the answers... weird.
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05 Jul 2005, 10:06
HowManyToGo wrote:
I got D.

a must be a multiple of 4. hence (a + 2)/2 = (4k + 2)/2 = 2k + 1 which is always odd.

HMTG.

very good explanation. Thanks HMTG.

D hits the point of the question "which of the following must be an odd integer ?? "

E is ruled out since (b+2)/2 is even in some situations, try b = 2, 6, 22, 26, 32, 36.....to any value ended up with last digit of 2 or 6)
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05 Jul 2005, 14:21
aowaowb wrote:
HowManyToGo wrote:
I got D.

a must be a multiple of 4. hence (a + 2)/2 = (4k + 2)/2 = 2k + 1 which is always odd.

HMTG.

very good explanation. Thanks HMTG.

D hits the point of the question "which of the following must be an odd integer ?? "

E is ruled out since (b+2)/2 is even in some situations, try b = 2, 6, 22, 26, 32, 36.....to any value ended up with last digit of 2 or 6)

plz do you have The OA
one thing for sure is that a and b must be either both even or both odd
so if you pick a=2 IN answer d (a+2)/2= 4/2 =2 this is a tricky question

'
05 Jul 2005, 14:21
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