1. The number $N$ has the smallest positive divisor 1, the second largest positive divisor $k$, and the third largest positive divisor $m$. Given that $k^{k}+m^{m}=N$. What is $N$?
Solution. Answer: $N=260$. If $N$ were odd, then both $k$ and $m$ would be odd, which means $k^{k}+m^{m}$ would be an even number. Therefore, the equation $N=k^{k}+m^{m}$ would be impossible in this case.

Now let $N$ be even. Then $k=2$ and $m$ is also even. Since $4=N-m^{m}$ is divisible by $m$, we have $m=4$ (since $m>k$). Therefore, $N=2^{2}+4^{4}=260$.