# Positive Integers Equal to Sum of Digits of Cube

## Theorem

The only positive integers which are equal to the sum of the digits of their cube are:

$0, 1, 8, 17, 18, 26, 27$

two of which are themselves cubes, and one of which is prime.

## Proof

We have trivially that:

 $\displaystyle 0^3$ $=$ $\displaystyle 0$ $\displaystyle 1^3$ $=$ $\displaystyle 1$

Then:

 $\displaystyle 8^3$ $=$ $\displaystyle 512$ $\displaystyle 8$ $=$ $\displaystyle 5 + 1 + 2$

 $\displaystyle 17^3$ $=$ $\displaystyle 4913$ $\displaystyle 17$ $=$ $\displaystyle 4 + 9 + 1 + 3$

 $\displaystyle 18^3$ $=$ $\displaystyle 5832$ $\displaystyle 18$ $=$ $\displaystyle 5 + 8 + 3 + 2$

 $\displaystyle 26^3$ $=$ $\displaystyle 17576$ $\displaystyle 26$ $=$ $\displaystyle 1 + 7 + 5 + 7 + 6$

 $\displaystyle 27^3$ $=$ $\displaystyle 19683$ $\displaystyle 27$ $=$ $\displaystyle 1 + 9 + 6 + 8 + 3$

A quick empirical test shows that when $n = 46$, it is already too large to be the sum of the digits of its cube.

## Also reported as

Some sources (either deliberately or by oversight) do not include $0$ in this list.

## Historical Note

David Wells report in his $1997$ work Curious and Interesting Numbers, 2nd ed. that this result can be found in an article by Monte James Zerger in Journal of Recreational Mathematics, volume $25$, page $248$, but this has not been corroborated, through want of evidence.

N.J.A. Sloane reports in A046459 of the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008) that the result originated with Claude Séraphin Moret-Blanc in $1879$.