\(x > 0\) and \(y > 0\) and \(xy + 5x + 2y + 9 = even\)

\(2y = even\)
\(9 = odd\)

So, \(xy + 5x = even\)
As, \(even + even + odd = even\)

Now, if \(xy\) is even then \(x\) has to even and \(y\) has to be odd

Now,
Col. A: Remainder greater than 0
Col. B: Remainder 0

Hence, option A

i attacked this question in this particular way.... i managed 53 seconds in total.

so xy + 5x +2y + 9 needs to be even and x and y are positive integers .

i would use random positive values for x and y both here
1) x=2 ,y=1 to keep everything even. but 9 here makes it odd. so then one and only ONE of the values needs to be odd for it to equal in total an even value
2) x= 3, y=2... why these two values .. i want my xy to be even. only 5x will end up as odd and my 2y remains even and that 9 value makes the odd value even at the end because odd+odd= even.

now that we have defined our variables we can move to the comparison for divison .

A = reamainder is 1 , B = remainder is 0

A is greater hence Option A

How I approached:

\(xy+5x+2y+9=even\)
\(xy+5x+2y=even-9=odd\)
\(xy+5x=odd-2y\)
\(xy+5x=odd\)
Knowing the above:
Notice that if x were even, the equation is invalid - as xy would be an even quantity, and 5x would also be even, and even plus even does not equal an odd value. Therefore, X must not be even.
We can deduce that quantity A is 1.

If x is odd, y may be even - xy=even, 5x=odd, even plus odd is odd.
In this case, quantity A is 1 and quantity B is 0. QA>QB

If x is odd, and y is odd - this is not possible, as xy would be odd and 5x would be odd, and odd+odd is even.

Therefore, we have only the case that x is an odd number, and y is an even number.

QA>QB, so the answer is A.

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