N equals the cube of the sum of its digits

[myth_kill]

Find all natural numbers n such that n equals the cube of the sum of its digits.

 

[jobsism]

If it were the sum of the individual cubes of the digits, then there is

1^3 +5^3 +3^3 =153

But I guess that's not the answer you require.

 

[sennyk]

I believe that there are 6 or 7 answers; n = 0, 1, 512, 4913, 5832, 17576, 19683. It depends if you are counting zero or not.

Proving this is another story. First, candidates for n must be equal to 'k' cubed. So we can eliminate a lot numbers with that statement. Once 'k' gets to be 54, (n = 157464), it is impossible for the sum of digits of n to be equal to its cube root (k). Thus one can perform an exhaustive search for k = 0..54 to find the only answers. Openoffice helps with the exhaustive search.

Am I missing something here? Any comments are appreciated.