In honor of David Hilbert's birthday, January 23

Question

Here is a skeletal division to celebrate David Hilbert's birthday.

Answer 1

We'll assume the normal rules of alphametics - that each letter corresponds to a different digit and we don't have leading zeroes

First note that

The highest multiple of $23$ less than $HIL$ has two digits, which means $H=1$. Also, this highest multiple must be $92$ and so, $D=4$ and $HIL$ cannot be greater than $115$. Hence $I=0$.

The division implies that $ID \times 23 + 1 = RT$ and since $ID$ is $04$, it must be that $RT$ is $93$.

Now, note that $V$ is a two-digit multiple of $23$, and since $V$ cannot be $1,3$ or $4$ (already taken), it must be that $V=2$. This puts $E=6$ and $A,L,B$ must be $5,7,8$ in some order.

A simple check shows that $A$ cannot be $8$ (otherwise $I=1$) and $A$ cannot be $5$ (otherwise $B=9$). Hence, $A=7$, $L=8$, $B=5$ and the division looks as follows

Summary

$DAVID = 47204$
$HILBERT = 1085693$