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Please help regarding the below question wid explanation.

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

Please help regarding the below question wid explanation.

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.

Answer A.
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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, 09:12

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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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14 Mar 2017, 10:16

Hello from the GMAT Club BumpBot!

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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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03 Apr 2017, 04:39

ab^2/c is an even integer. When is that possible: When b^2 or b is even (If b is even then b^2 would be even i.e. even*even is even. For example, 2*2=4)

a and c can either be even or odd. Why? Because an odd number can have an even multiple (e.g. 12 is multiple of 3). Also in 12, the first digit is odd and then also the number is even because it ends with an even digit.

Check the options: (1) ab is even. YES!! Because we know b is even. So odd*even=even (e.g. 3*2=6) or even*even=even (e.g. 2*2=4) (2) ab>0. Well!! It can be or can't be. Because ab^2/c is a positive integer. That tells us a few things: - If b is negative, then b^2 is positive. And a and c are positive. In this case, ab will be less than 0. - If a is negative and b is also negative, then ab>0 (negative*negative= positive). In this case ab>0. (Can eliminate this choice here only as we know that we are not getting a unique answer). - If b is not negative and c and a are negative, then ab will be less than zero.

So no unique answer.

(3) c is even. We already know that c can be odd or even. And in mUST BE TRUE questions, there SHOULD be ONLY one answer.

The answer is statement 1.
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If a, b, and c are integers and ab^2/c is a positive even in
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03 Apr 2017, 04:39

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