, 2025

$H$ is a perfect square and is $1$ more than $D$. The perfect squares from $1$ to $8$ inclusive are $1$ and $4$. Since $H$ is one more than $D$, $H$ cannot be equal to $1$ (since $D\neq0$), and so $H=4$ and $D=3$. $B$ is the largest prime number in the set, and so $B=7$. $C$ is a multiple of both $G$ and $D$. Since $D=3$, then $C=6$ and $G$ is equal to either $1$ or $2$ (since $6$ is a multiple of each of these). The value of $B+G$ is even, and since $B=7$, then $G=1$. The letters which have not been assigned values are $A$, $E$ and $F$, and the integers which have not been assigned are $2$, $5$ and $8$. Since $5$ and $8$ are in the same row and $A$ and $E$ are the two remaining unassigned letters that are in the same row, then $A$ and $E$ are $5$ and $8$ in some order, which leaves $F=2$.