If a, b, and c are integers and a*b^2/c is a positive even

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.

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.

Answer: A.

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We are given that (a*b^2)/c is a positive even integer. Therefore, a*b^2 must be even. (If a*b^2 is odd, (a*b^2)/c can’t ever be even.)

Now recall that the product of an even number and any integer is even, so either a or b, or both, must be even. Thus we see that ab must be an even integer. However, ab DOES NOT have to be greater than zero, since a could be -2 and b could be 1. Finally, we see that c does not have to be even, since a could be -2, b could be 1, and c = -1. Thus, only Roman numeral I must be true.

Answer: A

Can be solved less than 1 min conceptually!

We know that (these formulas can come handy to save time):

Case 1: even/even = even or odd
Case 2: even/odd = even

I. ab is even: in both cases, numerator needs to be EVEN, so this is a MUST be TRUE condition
II. ab>0: Not necessarily, to produce positive result, we need either both positive or both negative as numerator and denominator, so this does not need to be true always
II. c is even: From Case 1 and Case 2, it can be seen that the denominator can be even or odd. So, c can be either even or odd, so not true always!

Answer A. I only!

Given: \(a*b^2/c\) is an even integer

It means the numerator ab^2 is even i.e. either a or b or both must have 2 as a factor. Hence ab certainly has 2 as a factor and is even.
Hence, I must be true.

Given: \(a*b^2/c\) is positive
b^2 will be positive in any case - whether b is positive or negative.
This means a and c are either both positive or both negative.
So ab may be positive or negative since a can be positive or negative and b can be positive or negative. Hence ab doesn't need to be positive.

c needn't be even. c must be a factor of ab^2 but it may or may not have 2 as a factor.
e.g. c = 3, a = 6, b = 1
Here ab^2/c will be even integer though c is odd.

Answer (A)

Bunuel, I have a question on II. Why are we not considering the scenario of ab=0. That's when I thought II is "must be true". The question says nothing about a,b,c being != 0. 0 is an integer as well. Could you please explain? Thank you!

Responding to a pm:

Engineer1

Statement 2 says ab > 0

What that means is that when we multiply a and b, we get a positive number. Is it possible that even one of a and b is 0?
The moment one of them becomes 0, ab becomes 0 and cannot be positive.

KarishmaB, thanks for the quick response.

Yes, for this exact reason ab>0 is a must be true statement for ab (or ab^2) to be a positive integer, is what I was referring to. Therefore the answer should be I & II (C).

Think about it: Is it necessary that ab > 0 ? You are right that ab cannot be 0 but can't we have ab < 0?

Say a = 4, b = -1 and c = 2 (all integers while ab < 0)
Then
\(\frac{a*b^2}{c} = \frac{4 * (-1)^2}{2} = 2\) (positive even integer)

So ab < 0 is also possible. It is NOT NECESSARY that ab > 0

Cases:

All numbers can be negative
b can be negative (a & c positive)
a & c can be -ve

Even/Odd = Even and Even/Even= Even

ab^2/c= Even
ab^2= C*Even = Even
ab^2=Even
Therefore, ab has to be even

ab can be less than zero since (from above) b can be negative and a can be positive - Hence ab>0 is incorrect

c can be even or odd since,

even/odd = even and
even/even = even

since ab^2 is even c can either be even or odd. Therefore 'c is even' is incorrect

IMO A

Please correct logic wherever required. Thanks