Today a question about number theory just popped out in my mind.
Consider all the the permutations of one $1$, two $2$'s, three $3$'s,..... , nine $9$'s. Can any of its permutation be ever expressed in the form of $a^b$ such that $a$ and $b$ are both natural numbers. I want to find the least value of $b$ for which at least one of the permutation can be expressed in form $a^b$.
I have tried up-to $13^{th}$ powers using the residue mod $9$ but none of them was $\equiv 6\pmod 9$. Can someone please share some thoughts over it.
Heuristically, it looks promising. There are $$\binom{45}{1,2,3,4,5,6,7,8,9} \approx 10^{34.8}$$ numbers with that digit distribution and approximately $1$ in $10^{22.5}$ should be a square, and $1$ in $10^{30}$ should be a cube.
However, you make a good observation about values modulo $9$.
$1^1 + 2^2 + \ldots + 9^2 \equiv 6 \pmod 9$, and $a^b \equiv 6 \pmod 9$ clearly has no integer solutions for $b > 1$ since $a$ would clearly have to be a multiple of $3$, but then $a^b$ is a multiple of $9$.