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\begin{document}
\indent For any polynomial \ $F(x)$, let \ $\bar{F}$ \ denote the
reduction (to the least nonnegative residue) (\mbox{mod} \ 3)

\noindent Question: \ \ If \ $F(x) \ = \ (1 + x)^n, \
                        \mbox{find} \ \bar{F}(1)$

\indent Write \ $ n = \displaystyle{\sum^J_{j=1}} a_j e^j$


Let \ $A \ = \ \{j: \ a_j=1\}; \ \ B = \{j: \ a_j=2\} \\
\ \ \ \ |A| = a; \ \ \ |B| = b$.

\indent Let \ $G(x) = \
                \displaystyle{\pi^J_{j=1}} \ (1+x^{3^j})^{a_j}$

\indent Since \ $(1 + x)^3 \ \equiv \ 1 + x^{3^j} \ (\mbox{mod} \ 3)$, \
 we have \ $G(x) \ \equiv \ F(x) \ (\mbox{mod} \ 3)$.

\indent Hence the problem reduces to finding \ $\bar{G}(1)$.  The advantage in
\ $G$ \ is that there is no carry over, that is every monomial in the
product occurs exactly once.

\indent Now, \ $G(x) = \displaystyle{\Pi_{j\in B}} (1 + 2x^{3^j} +
x^{2 \star 3^j})  \displaystyle{\Pi_{j \in A}} \left(1 + x^{3^j}\right) \ =
    \ \sum b_j x^j$

$$\begin{array}{lccl}
    \mbox{Let} & R & = & |\{j: b_j \equiv 2 (\mbox{mod} \ 3)\} \\
               & S & = & \{j: \ b_j \equiv 1 (\mbox{mod} \ 3)\} \end{array}
$$

\indent so that \ $\bar{G}(1) \ = \ S + 2R$.

\indent From the expansion of \ $G(x)$, \ it is clear that if each
non zero \ $b_j$, is replaced by \ $1$, we get

$$
\displaystyle{H(x) = \Pi_{j\in B}} \ (1 + x^{3^j} + x^{2
\star 3^j})
\ \displaystyle{\Pi_{j\in A}} \ \left(1 + x^{3^j}\right)
$$

\noindent and hence \ $R + S = H(1) = 3^b 2^a$.

\indent Also if we replace \ $b_j$ \ by \ $(-1)$ \ for \ $j \in R$, \
and \ $b_j \ \mbox{by} \ (+1)$ \ for \ $j \in S$, \ we end up with

$$
\displaystyle{\Pi_{j\in B}} \left(1 - x^{3^j} + x^{2 \star 3^j}\right) \
\displaystyle{\Pi_{j\in A}} \left(1 + x^{3^j}\right)
$$

\noindent and hence \ $S - R \ = \ 2^a$.

\indent Thus \ $S \ = \ 2^{a-1} (3^b + 1)$ \ and \
$R \ = \ 2^{a-1} (3^b - 1)$.

\indent Hence \ $\bar{G}(1) \ = \ S + 2R \ = \ 2^{a-1} (3^{b+1} - 1)$
\end{document}