The difference between '$KAN$' and '$GAR$' is less than $100$ and since $K\neq G$ we must have $K=G+1$.
Next we must have $N < R$ else the difference between '$KAN$' and '$GAR$' would be at least $100$.
Let $R=N+x$ where $1< x< 9$. Then $OO=100-x$ and hence $O=9$ and $R=N+1$. Also we must have $K\leq 8$
We want the largest value for $KAN$ so we try $K=8$. This forces $G=7$, hence must have $A\leq 6$. Set $A=6$, this forces $R\leq 5$ and hence $N\leq 4$ since $R=N+1$.
So $864$ is the largest possible value for $KAN$, and we have $864-765=99$.
This problem is taken from the UKMT Mathematical Challenges.