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 $864-765=99$.

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*This problem is taken from the UKMT Mathematical Challenges.**View the archive of all weekly problems grouped by curriculum topic*