\documentclass[12pt]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{amsfonts}
\usepackage{geometry}
\usepackage{hyperref}

%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Version=5.00.0.2570}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{Created=Tuesday, September 21, 2004 06:51:08}
%TCIDATA{LastRevised=Sunday, January 08, 2006 14:38:09}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
%TCIDATA{CSTFile=40 LaTeX article.cst}

\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}
\geometry{left=0.6in,right=0.5in,top=0.4in,bottom=0.5in}

\begin{document}


\begin{center}
{\Large Puzzle 6 - SOLUTION}

Not Mindreading
\end{center}

\bigskip

\begin{problem}
Instructions:

1. Think of a 5-6 digit number, that has at least two different digits in
it. My example is $803225$.

2. Create a second number by rearranging the digits of the previous number.
My example is $320258$.

3. Subtract the smaller number from the larger number. My example is $%
803225-320258=\allowbreak 482\,967$

4. Cross out any non-zero digit of the difference. My example is $482\,9\NEG%
{6}7$.

5. Announce the number you obtain by omitting the crossed out digit. My
example is $482\,97$.

If you tell the mindreader the last number, s/he can tell what digit you've
crossed out. How?
\end{problem}

\begin{solution}
The trick is based on the rule of divisibility by $9$. The rule is that
every number's reminder after division by $9$ is the same remainder if we
divide the sum of its digits by $9$. For example, the sum of the digits in
the number $2500041$ is $2+5+4+1=12$. When we divide $12$ by $9$ we get a
remainder $3$. This means that if we divided $2500041$ by $9$, we get a
remainder $3$. In particular, if the sum of the digits is divisible by $9$,
so is the number.

This trick expolits this property. When we create the second number, we use
the same digits. So, we obtain two numbers that have the same sum of digits
and thus the same remainder when divided by $9$. This means that the third
number, the difference of the frist two, will certainly be divisible by $9$
and so its digits add up to a number divisible by $9$.

When someone announces their final number, we need to add the digits and see
what number needs to be added to it to obtain something divisible by $9$.
(Now you see why crossing out $0$ was not allowed, we wouldn't be able to
distinguish it from $9$.)

In my example, we hear $482\,97$. Since $4+8+2+9+7=\allowbreak 30$, we need
add $6$ to that to make it divisible by $9$. (The next number, $15$ is too
large). Thus they crossed out a $6$.
\end{solution}

\end{document}