Statement I. ab is even

GIVEN: \(\frac{ab^2}{c}\) is an even integer

This means we can say that \(\frac{ab^2}{c}\) = 2k (for some integer k)

Multiply both sides by c to get: \(ab^2 = 2kc\)

We can see that 2kc must be EVEN, which means ab^2 must be EVEN.
If ab^2 is EVEN, then either a or b must be EVEN, which means ab must be EVEN
So statement I is true

Check the answer choices....ELIMINATE B
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Statement II. ab > 0
Notice that, regardless of the value of b, we know that b² is POSITIVE (for all non-zero values of b)
This leads me to test some possible values...

If \(\frac{ab^2}{c}\) is a positive even integer, then it COULD be the case that a = 2, b = -1 and c = 1
Notice that \(\frac{ab^2}{c}=\frac{(2)(-1)^2}{1}=2\), which is a positive even integer

In this case, ab = (2)(-1) = -2
So, it is NOT true that ab > 0
So statement II is NOT true

Check the answer choices....ELIMINATE C and E
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Statement III. c is even
Notice that we can reuse the values we used above (a = 2, b = -1 and c = 1)
If c = 1, then c is NOT even
So statement III is NOT true

Check the answer choices....ELIMINATE D

The correct answer is A

Cheers,
Brent