I have checked the question upon this topic, but everytime the answer is the parity of zero is even because $0\times2=0$, and it is between two odd numbers, $-1,1$.

My question is, $0\times\text{(any odd number)}$ is also equal to zero, making the odd number a factor of zero, and thus $0$ could also be said to be odd.

Also, the logic that every even number lies between two odd numbers may be an exception for zero, as such exceptions occurs frequently in Number Theory.

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  • 1
    $\begingroup$ I don't understand your doubt. $3$ is odd, $6$ is even and $3$ is a factor of $6$. It happens all the time, really. $\endgroup$
    – user228113
    May 16, 2017 at 8:54
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    $\begingroup$ According to your logic, 3 (an odd number) is also a factor of 6 (an even number) so 6 can also be said to be odd. This is of course not the case. While odd numbers only have odd divisors, it does not mean that only odd numbers have odd divisors. $\endgroup$ May 16, 2017 at 8:55
  • $\begingroup$ I think you have a very odd approach to thinking about the parity of a number. Whether the number has odd factors, has nothing to do with it. Solution? Just take the number mod 2; 0 mod 2 = 0, making 0 an even number. $\endgroup$
    – Matti P.
    May 16, 2017 at 8:58
  • $\begingroup$ The odd number $1$ is also a factor of $2$, so by your logic, is $2$ also odd? $\endgroup$
    – 5xum
    May 16, 2017 at 9:00

3 Answers 3


An even number, say $k$, is a number where $$k\mod 2=0$$

Therefore, we set $k=0$ and note that $$0\mod 2=0$$ and therefore $0$ is even

We could also say that we know $1$ is odd, and therefore $$1\mod 2=1$$

We can note that $0=1-1$ and therefore \begin{align}0\mod 2&\equiv 1-1\mod 2\\ &\equiv (1\mod 2) - (1\mod 2)\\ &\equiv 1-1\\ &\equiv 0\end{align}

There is also a whole Wikiepdia page dedicated to this problem!


An even number times an odd number is always even. For example, $2\cdot 3 = 6$. $3$ is a factor of $6$, but $6$ is still even. Being a multiple of an odd number doesn't make a number odd.

It's simply true by definition that $k\in \mathbb{Z}$ is even if there exists $n\in \mathbb{Z}$ such that $k=2n$. This applies to $0$, and so $0$ is an even number.


You've got things backwards. If $a$ is a factor of $b$ and $b$ is odd, then $a$ is odd (e.g. since $9$ is odd and $3$ divides $9$, we know that $3$ is odd - although this is a bit of a silly way to find that out!). However, odd numbers can be factors of even numbers without a problem.

More to the point, by definition a number is even if it is of the form $2k$ for some integer $k$. Since $0=2\cdot 0$, this means $0$ is even.


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