The exact question is here
A whole number is equal to the arithmetic mean of all the numbers obtained from the given number with the aid of all possible permutation of its digits. Find all whole numbers with that property.
Answer
Take $k=a_n\cdot10^{n-1}+a_{n-1}\cdot10^{n-2}+\dots+a_2\cdot10^1+a_1\cdot10^0$
The sum of all the possible permutation of k is:
$$\frac{(n-1)!}{m}\cdot(a_n+a_{n-1}+\dots+a_1)\cdot\underbrace{111\dots111}_{n\text{ 1's}}$$
With $m$ being the divisor if any of $a_n, a_{n-1}, \dots , a_1$ are the same.
The average of all the possible permutation of k would be:
$$\frac{\frac{(n-1)!}{m}\cdot(a_n+a_{n-1}+\dots+a_1)\cdot\underbrace{111\dots111}_{n\text{ 1's}}}{\frac{n!}{m}}\\=\frac{\underbrace{111\dots111}_{n\text{ 1's}}\cdot(a_n+a_{n-1}+\dots+a_1)}{n}\\=k=a_n\cdot10^{n-1}+a_{n-1}\cdot10^{n-2}+\dots+a_2\cdot10^1+a_1\cdot10^0
$$
Now i got the mean, however I'm not sure how to continue.
Can someone help me with this?
Thank you.