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alphametic – In honor of David Hilbert’s birthday, January 23

We’ll assume the conventional guidelines of alphametics – that every letter corresponds to a distinct digit and we do not have main zeroes

First word that

The very best a number of of $23$ lower than $HIL$ has two digits, which suggests $H=1$. Additionally, this highest a number of have to be $92$ and so, $D=4$ and $HIL$ can’t be larger than $115$. Therefore $I=0$.

The division implies that $ID instances 23 + 1 = RT$ and since $ID$ is $04$, it have to be that $RT$ is $93$.

Now, word that $V$ is a two-digit a number of of $23$, and since $V$ can’t be $1,3$ or $4$ (already taken), it have to be that $V=2$. This places $E=6$ and $A,L,B$ have to be $5,7,8$ in some order.

A easy verify exhibits that $A$ can’t be $8$ (in any other case $I=1$) and $A$ can’t be $5$ (in any other case $B=9$). Therefore, $A=7$, $L=8$, $B=5$ and the division seems to be as follows
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$DAVID = 47204$
$HILBERT = 1085693$



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