Given: \(x(x + 3) – 153 = xy\)
Expand: \(x^2 + 3x – 153 = xy\)
Add \(153\) to both sides of the equation: \(x^2 + 3x = xy + 153\)
Subtract \(xy\) from both sides of the equation: \(x^2 + 3x - xy = 153\)
Factor the left side: \(x(x + 3 - y) = 153\)

At this point it's crucial to recognize that x and y are INTEGERS. So, the above equation can be expressed as follows: (some integer)(some other integer) = some ODD integer.
If the product of two integers is ODD, then both of those integers must be ODD.
In other words, x is ODD and (x + 3 - y) is ODD

Since x is ODD, we can replace x in (x + 3 - y) to conclude that (odd + 3 - y) is ODD.
Since 3 is also odd, we can write: (odd + odd - y) = ODD, which means y must be ODD

Since x and y are both ODD, both values yield a remainder of 1 when divided by 2.

So we have:
QUANTITY A: 1
QUANTITY B: 1

Answer: C