A number theory problem by Kenny O.

We know that for $n^2$ and $n^3$ to have all the digits, the number of the digits in each number must add up to 10.
We also know that $n^2$ would have an equal or less amount of digits than $n^3$ .
If $n^2$ has 1 digit, $n^3$ would need to have 9 digits but the most digits for $n^3$ , if $n^2$ has only 1 digit, is 2 (27)
If $n^2$ has 2 digits, $n^3$ would need to have 8 digits but the most digits for $n^3$ , if $n^2$ has 2 digits, is 3 (729)
If $n^2$ has 3 digits, $n^3$ would need to have 7 digits but the most digits for $n^3$ , if $n^2$ has 3 digits, is 5 (29791)
If $n^2$ has 4 digits, $n^3$ would need to have 6 digits, which is possible. The range is 48<n<100.
For all the digits to appear exactly once, the units digits cannot be 1, 5, 6 and 0.
The sum of all numbers is 45, the sum of $n^2$ and $n^3$ must be a multiple of 9. $n^2(n+1)=9k$ where k is an integer. Thus, n is a multiple of 3 or n=9a+8 where a is an integer.
This reduces the possible answers to:
53, 54, 57, 62, 63, 69, 72, 78, 84, 87, 89, 93, 98 and 99.
Doing manual calculation, you will get 69.
$69^2$ = 4761
$69^3$ = 328509