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# If a, b, and c are integers and ab^2/c is a positive even in

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If a, b, and c are integers and ab^2/c is a positive even in [#permalink]

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19 Jun 2013, 23:04
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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

A. I only
B. II only
C. I and II
D. I and III
E. I, II, and III
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Jun 2013, 23:50, edited 1 time in total.
Renamed the topic, edited the question and the tags.
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19 Jun 2013, 23:14
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Countdown wrote:

If a , b, and c are integers and ab2 / 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

ab2/c is positive which means a and C >0 or a and c<0 since b2 is positive
Case 1 - a=3 b=2 c=3 or c=6 where ab2 /2 would be even when c= 3 or c=6
lets see III option where C is even as per our assumption C can be even and odd, hence it is out.
lets see option II ab>0 a and b can positive or negative hence sign of ab cannot be determined hence it is out

So options B,C,D,E are gone and ans is A

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19 Jun 2013, 23:26
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Countdown wrote:

If a , b, and c are integers and ab2 / 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

I believe the expression is $$ab^2/c$$ is a positive integer.
Now we know,

1. even * even = even (2*4 = 8)
2. even * odd = even (2*3 = 6)
3. even / even = even or odd (8/2 = 4 , 6/2 = 3)
4. even / odd = even (6/3 = 2)
5. odd * odd = odd
6. odd / odd = odd (if the are divisible)
7. odd / even = <not divisible> as the denominator will always have an extra 2.

Coming back to my question, my result is even. Hence my numerator must be even, as odd numerator can never give even result. (we are considering divisible integers only)

$$ab^2$$ is even, hence either a is even or $$b^2$$ is even. If $$b^2$$ is even then b is even. Hence ab = even. (I) is true.
now $$\sqrt{b^2}$$ = +/- b, so ab can be > 0 or < 0, (II) is false
From rule 3 and 4, c can be even or odd. so (III) is false.

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Re: If a, b, and c are integers and ab^2/c is a positive even in [#permalink]

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20 Jun 2013, 00:11
Countdown 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

A. I only
B. II only
C. I and II
D. I and III
E. I, II, and III

Given: a, b, and c are integers and ab^2/c is a positive even integer.

Evaluate each option:

I. ab is even.

ab^2/c to be positive even integer ab^2 must be even, from which it follows that either a or b must be even, thus ab has to be even.

II. ab > 0

Since b is squared in our expression, then we can say nothing about its sign, thus we cannot say whether ab is positive.

III. c is even

The easiest one: ab^2/c = integer/c = even --> c could be even as well as odd.

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Re: If a, b, and c are integers and ab^2/c is a positive even in [#permalink]

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21 Oct 2014, 08:12
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Re: If a, b, and c are integers and ab^2/c is a positive even in   [#permalink] 21 Oct 2014, 08:12
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