From a list of integers from $1$ to $2022$, inclusive, delete all numbers in which at least one of its digits is a prime How many numbers remain without erasing?
We will count the numbers from $0$ to $2022$ not having any primes among its digits, erasing $0$ at the end.
There are $6$ one-digit numbers fulfilling the condition: $0$, $1$, $4$, $6$, $8$, $9$.
Now, for two-digit numbers, we have that the first digit must be one of the above without $0$ and the second must be one of the above, so we have $5 \cdot 6=30$ two-digit numbers fulfilling the condition.
For three-digit numbers, we similarly get $5 \cdot 6 \cdot 6=180$ numbers fulfilling the condition.
For four-digit numbers, the first digit can only be $1$ or $2$. As $2$ is not possible given the condition, all these numbers must start with $1$. After that, there are similarly $6 \cdot 6 \cdot 6=216$ possibilities.
Obviously, all mentioned numbers are in $[0;2022]$ and distinct. So in total, without $0$, there are $6+30+180+216-1=431$ possible numbers. As these are the numbers not being erased, the answer is $\boxed{431}$.