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?

Question

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

Answer 1

The choice of 0 is incorrect, because if the product of integers is even, at least one must be even. (The product of (four) odd numbers is odd.) In your example, A=0, B=2, C=0, D=2, the four numbers are even (as is their sum and product). Also given in the comments is an example where two of the numbers are even (as is their sum and product), viz. A=0; B=1; C=0; D=3.

Answer 2

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.