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%%% Primary Title: Zuckerman number
%%% Primary Category Code: 11A63
%%% Filename: ZuckermanNumber.tex
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%%% Owner: CompositeFan
%%% Author(s): CompositeFan
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\begin{document}

 Consider the \htmladdnormallink{integer}{http://planetmath.org/encyclopedia/RationalInteger.html} 384. Multiplying its \htmladdnormallink{digits}{http://planetmath.org/encyclopedia/PlaceSystems.html}, $$3 \times 8 \times 4 = 96$$ and $${{384} \over {96}} = 91.$$

When an integer is \htmladdnormallink{divisible}{http://planetmath.org/encyclopedia/DivisibilityOfIdeals.html} by the \htmladdnormallink{product}{http://planetmath.org/encyclopedia/Product.html} of its digits, it's called a {\em Zuckerman number}. That is, given $m$ is the \htmladdnormallink{number}{http://planetmath.org/encyclopedia/Number.html} of digits of $n$ and $d_x$ (for $x \le k$) is an integer of $n$,

$${\prod_{i = 1}^m d_i}|n$$

All 1-digit numbers and the \htmladdnormallink{base}{http://planetmath.org/encyclopedia/PlaceSystems.html} number itself are Zuckerman numbers.

It is possible for an integer to be divisible by its \htmladdnormallink{multiplicative digital root}{http://planetmath.org/encyclopedia/MultiplicativeDigitalRoot.html} and yet not be a Zuckerman number because it doesn't \htmladdnormallink{divide}{http://planetmath.org/encyclopedia/Multiple2.html} its first digit product evenly (for example, 1728 in base 10 has multiplicative digital root 2 but is not divisible by $1 \times 7 \times 2 \times 8 = 112$). The reverse is also possible (for example, 384 is divisible by 96, as shown above, but clearly not by its multiplicative digital root 0).

\begin{thebibliography}{2}
\bibitem{jt} J. J. Tattersall, {\it Elementary number theory in nine chapters}, p. 86. Cambridge: Cambridge University Press (2005)
\end{thebibliography}


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