# Sequence of Dudeney Numbers

## Theorem

The only Dudeney numbers 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:

 $\ds 0^3$ $=$ $\ds 0$ $\ds 1^3$ $=$ $\ds 1$

Then:

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

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

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

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

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

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

For $46 < n \le 54$, $n^3 \le 54^3 < 200 \, 000$.

Hence the sum of the digits of $n^3$ is less than:

$1 + 5 \times 9 = 46 < n$

For $54 < n < 100$, $n^3 < 10^6$.

Hence the sum of the digits of $n^3$ is less than:

$6 \times 9 = 54 < n$

For $n \ge 100$, let $n$ be a $d$-digit number, where $d \ge 3$.

Then $10^{d - 1} \le n < 10^d$ and $n^3 < 10^{3 d}$.

Hence the sum of the digits of $n^3$ is less than:

 $\ds 3 d \times 9$ $=$ $\ds 27 d$ $\ds$ $\le$ $\ds 27 d + 63 d - 189$ $\ds$ $<$ $\ds 90 d - 170$ $\ds$ $=$ $\ds 10 \paren {1 + 9 \paren {d - 2} }$ $\ds$ $\le$ $\ds 10 \times \paren {1 + 9}^{d - 2}$ Bernoulli's Inequality $\ds$ $\le$ $\ds n$

so no numbers greater than $46$ can have this property.

$\blacksquare$

## Also reported as

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

## Historical Note

The earliest known appearance of this result is from Claude Séraphin Moret-Blanc in $1879$, although exactly where this was published is still to be identified.

Henry Ernest Dudeney subsequently published it in one of his own collections.

As a result, a number which equals the sum of the digits of its cube is now called a Dudeney number.

It continues to crop up occasionally in publications devoted to recreational mathematics.