Narcissistic numbers in other bases It is well known that $153$ is a narcissistic number; that is, it is equal to the sum of the cubes of its digits since $153=1^3+5^3+3^3$.
Other bases have similar numbers.  For example, in base $3$, seventeen is $122$; and in base $4$, thirty-five is $203$.
Let $B_3$ be the set of bases with no such [edit] three-digit numbers.  The first two members of $B_3$ are $2$ and $72$.
Why is every member of $B_3$ except $2$ a multiple of $9$?

Proof by magic:
$$
\begin{eqnarray}
(k+1)^3 & + & 0^3 & + & (2k+1)^3 & = & (3k+1)^2 (k+1) & + & (3k+1)0 & + & (2k+1) \\
k^3 & + & 0^3 & + & (2k+1)^3 & = & (3k+2)^2k & + & (3k+2)0 & + & (2k+1) \\
(5k+1)^3 & + & (4k+2)^3 & + & (6k+2)^3 & = & (9k+3)^2(5k+1) & + & (9k+3)(4k+2) & + & (6k+2) \\
(7k+5)^3 & + & (2k+1)^3 & + & (6k+4)^3 & = & (9k+6)^2(7k+5) & + & (9k+6)(2k+1) & + & (6k+4)\\
\end{eqnarray}
$$
The first two identities show that any base not divisible by $3$ admits at least one three-digit narcissistic number; with the sole exception of base $2$ (the number resulting from the second identity would be $001$; not a proper three-digit one). The other two lines cover bases which are multiples of three but not multiples of $9$; again showing that each of them admits at least one narcissistic number. Thus, only bases which are multiples of $9$ can possibly not admit any narcissistic number (well, other than base $2$, of course).
