9. Find the smallest positive value of $36^{\mathrm{k}}-5^{\mathrm{m}}$, where $\mathrm{k}$ and $\mathrm{m}$ are positive integers.
9. $36^{\wedge} \mathrm{k}=6^{\wedge}\{2 \mathrm{k}\}$ always ends in 6 , while $5^{\wedge} \mathrm{m}$ always ends in 5 .
So the difference $36^{\wedge} \mathrm{k}-5^{\wedge} \mathrm{m}$ always ends in 1 . But it cannot be equal to 1 .
Indeed, $6^{\wedge}\{2 k\}-1-5^{\wedge} \mathrm{m}=\left(6^{\wedge} \mathrm{k}-1\right)\left(6^{\wedge} \mathrm{k}+1\right)-5^{\wedge} \mathrm{m}=!=0$ since $5^{\wedge} \mathrm{m}$ is only divisible by 5 , but 5 does not divide $6^{\wedge} \mathrm{k}+1$.
Smallest value is 11 , when $\mathrm{k}=1$ and $\mathrm{m}=2$.