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Even-odd numbers

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Even-odd numbers [#permalink]

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New post 20 May 2009, 17:42
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If a , b, and c are integers and \(\frac{ab^2}{c}\) is a positive even integer, which of the following must
be true?


I. ab is even
II. ab > 0
III. c is even

I only
II only
I and II
I and III
I, II, and III

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Re: Even-odd numbers [#permalink]

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New post 20 May 2009, 18:32
II only

example

a=3, b=3, c=3

3(3^2)/3

ab= (3x3) is not even and c=(3) is not even

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Re: Even-odd numbers [#permalink]

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New post 20 May 2009, 18:46
If a=b=c=3 then ab^2/c will not be an even positive integer.

How do I choose the right numbers?

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Re: Even-odd numbers [#permalink]

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New post 20 May 2009, 19:43
A.

ab has to be even for ab^2/c to be even. Rest need not be true.

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Re: Even-odd numbers [#permalink]

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New post 26 May 2009, 00:11
c doesn't have to be even, as b can be 3 and c can be 3, and a is even.

ab>0, not necessarily true. we could have a<0 b>0 and c<0, in which case ab/c >0 but ab<0.

Only ab has to be even, because if it weren't, that means it doesn't have a multiple of 2, and since c is a factor of ab, neither does c.
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Re: Even-odd numbers [#permalink]

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New post 26 May 2009, 10:54
IMO- II only
Whats the OA?

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Re: Even-odd numbers [#permalink]

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New post 26 May 2009, 11:02
chill wrote:
IMO- II only
Whats the OA?



a=-1
b=1
c=-1

ab^2/c>0 but ab<0 ..........
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Re: Even-odd numbers [#permalink]

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New post 26 May 2009, 23:04
a,b,c are integers

(a * b * b) / c = +ve ( even)

From this we know that when the numerator is odd then we will not get a +ve ( even) integer.
So the numerator needs to be even.
So either a or b need to be even so ab is even.

Other two are not true.

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Re: Even-odd numbers [#permalink]

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New post 27 May 2009, 08:36
ab^2/c = ab*b/c

lets assume a =1, b=2 than 4/c in this case ab = 2 (even)

But if c = 4 than a*b*b/c = 1 & this is not +ve even number

So I only is not true.

ab >0 does not conclude anything, see the above example.

So II only is not true

Now come of 3rd statement c is even. Again in the above example c is even but still ab62/c is not even .

So III only is also not sufficient.

And also No Other combination of I, II & III .

For me this question is Inconclusive.

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Re: Even-odd numbers [#permalink]

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New post 27 May 2009, 08:57
gmatprep09 wrote:
If a , b, and c are integers and ab^2/c is a positive even integer, which of the following must be true?

I. ab is even
II. ab > 0
III. c is even

I only
II only
I and II
I and III
I, II, and III


ab^2/c = 2k where k is any +ve integer.

I. ab is even - true: Since 2k is even, either of a or b must be even. Any odd (if ab^2 is odd) divided by any integer (c must be odd) never results in even. So either a or b must be even.

II. ab > 0 - Not true: If a and c are +ve, and b is -ve, 2k is +ve and ab < 0.
If a, b and c, all, are -ve, 2k is still +ve and ab> 0.

III. c is even - Not true: If a = 5, b = 4 and c = 1, 2k is even. If a = 4, b = 4 and c = 4, 2k is even. So C is not necessarily an even.

Only I is correct.
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Re: Even-odd numbers   [#permalink] 27 May 2009, 08:57
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