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This is a follow-up to this post. In the post, the user defines a number $x$ that is $1$ if $\textsf{ZFC}$ is inconsistent and $0$ otherwise. By virtue of the fact that $x$ is either $0$ or $1$, it is computable. Now I am trying to compare this to the case of Chaitin's number $\Omega$, which is uncomputable.

Why exactly is $x$ computable and $\Omega$ is not? As someone in the link said,

More generally, you should be aware that not knowing what a number's value is has little to do with whether the number is computable.

This makes sense, but then why does this statement not apply to $\Omega$? What exactly am I missing that makes $x$ and $\Omega$ dissimilar?

Maximal Ideal
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    The difference is that $\Omega$ has an infinite number of digits. Any number $x$ with a finite number of digits is computable, because the algorithm to compute it is just "print(x)". – Jair Taylor Feb 25 '23 at 20:47

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