If both the product and sum of four integers are even, which of the following could be the number of even integers in the group? 0 2 4?

The answer says that only options 2 and 4 are right and option 0 is impossible.

But here is my solution which proves that option 0 is possible:

Let’s A=0, B=2, C=0, D=2;

So their sum is 0+2+0+2=4 (which is an even number)

And their product is 0*2*0*2 = 0 (which is also an even number)

So why choice 0 is incorrect?

Here is what the book tells me about it:

“Since these four integers have an even product, at least one of them must be even, so roman numeral I, 0, is impossible” . But I don't understand why

• The product of odd numbers is always odd. Your example would appear to have $4$ even integers. – lulu Feb 3 at 12:57
• In your example you have 4 even numbers, not $0$. – franzk Feb 3 at 13:00
• Maybe I don't understand your question. You give an example with $4$ even digits but then ask about the case with $0$ even digits. It's clear that you can't have no even digits as the product of odd digits is odd. – lulu Feb 3 at 13:00
• You can't have no even integers because the product of odd integers is odd. Your example with four even integers does not appear to be relevant. – lulu Feb 3 at 13:04
• I think you have failed to grasp the question you were asked. You aren't being asked if it is possible for any of the integers to be $0$, you are asked to count the number of even integers and decide whether that count might be $0$. – lulu Feb 3 at 13:14

It might help to look at this algebraically. Let's say $$a$$, $$b$$, $$c$$, $$d$$ are any integers whatsoever.
Then the integers $$2a + 1$$, $$2b + 1$$, $$2c + 1$$, and $$2d + 1$$ are all odd. Their sum is even: $$2a + 2b + 2c + 2d + 4$$. However, their product is $$16abcd + 8abc + 8ab +$$ yada, yada, yada, $$+ 2c + 2d + 1$$. You can have Wolfram Alpha do it for you if you don't feel like going through it yourself.
Now let's try $$2a + 1$$, $$2b + 1$$, $$2c + 1$$, and $$2d$$. There's only one even integer in here, but it's enough to make the product even. But now the sum is odd: $$2a + 2b + 2c + 2d + 3$$.
It's the same situation for $$2a + 1$$, $$2b$$, $$2c$$, and $$2d$$: even product but odd sum.