N equals the cube of the sum of its digits

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The discussion focuses on identifying natural numbers \( n \) that equal the cube of the sum of their digits. The confirmed solutions include \( n = 0, 1, 512, 4913, 5832, 17576, \) and \( 19683 \). The method involves eliminating candidates by establishing that \( n \) must be equal to \( k^3 \), where \( k \) represents the sum of the digits. An exhaustive search for \( k \) ranging from 0 to 54 is necessary, as beyond this range, the sum of the digits cannot equal \( k \). OpenOffice is suggested as a tool for conducting this exhaustive search.

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

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