1. For a natural number $n$, the smallest divisor $a$, different from 1, and the next largest divisor $b$ were taken. It turned out that $n=a^{a}+b^{b}$. Find $n$.
Answer: $n=2^{2}+4^{4}=260$.

Solution. If $n$ is odd, then all its divisors are also odd. Then $a$ and $b$ are odd and $a^{a}+b^{b}$ is even. Therefore, $n$ is even. Then its smallest divisor, different from 1, is 2, and thus $a=2$. Therefore, $n=2^{2}+b^{b}$. Consequently, $4=n-b^{b}$ is divisible by $b$. Then $b=4$, since $b>2$. Thus, $n=2^{2}+4^{4}=260$.