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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 $9$?

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  • $\begingroup$ This is a good place to start en.wikipedia.org/wiki/… $\endgroup$ Jun 21, 2013 at 2:54
  • $\begingroup$ A more concise way to say "in base 3, seventeen is 122" is “$122_3 = 17$”. $\endgroup$
    – MJD
    Jun 21, 2013 at 6:37
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    $\begingroup$ What is your definition of a narcissistic number? The one found in Wikipedia allows, for example, $305 = 4\times 72 + 17$ to be a narcissistic number in base $72$ (since $305 = 4^2 + 17^2$). Do you require the number to be three-digit in the chosen base or is there some additional constraint imposed? (of course, $0$ and $1$ are narcissistic numbers in all bases in the usual definition) $\endgroup$ Jun 21, 2013 at 6:43
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    $\begingroup$ @Michael How much numerical evidence do you have for this conjecture? $\endgroup$
    – Jim Belk
    Jun 22, 2013 at 3:02
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    $\begingroup$ The first few bases without a solution are 72,90,108,153,270,423,450,531,558,630,648,738,1044,1098,1125,1224,1242,1287,1440,1503,1566,1611,1620,1800,1935 $\endgroup$
    – Empy2
    Mar 21, 2014 at 7:18

1 Answer 1

9
+50
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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).

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