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\lhead{Mathematics and Digital Art}
\chead{}
\rhead{Due:  22 November 2017}
\lfoot{Matsko}
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\rfoot{17 November 2017}

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\begin{document}
We are working in the integers, denoted by ${\mathbb Z}.$

We say $a$ {\em divides} $b$ and write ``$a\mid b$'' if there is an integer $k$ such that $ka=b.$ 

If $a\mid b$ and $a\mid c,$ then $a\mid(b+c),$ $a\mid (b-c),$ and $a\mid (bc).$  We do not have a result for division (yet) since $a/b$ might not be an integer.

We say $a$ (mod $m$) $= r$ if $a=qm+r$ where $0\le r<m.$  $q,$ $r,$ are the quotient and remainder. $m$ is a positive integer at least $2.$

We write $a\equiv b$ (mod $m$) if $a$ (mod $m$) $= b$ (mod $m$), or equivalently, $m\mid (b-a).$

If $a\equiv b$ (mod $m$) and $c\equiv d$ (mod $m$), then:
\begin{enumerate}
\item $a+c\equiv b+d$ (mod $m$),
\item $a-c\equiv b-d$ (mod $m$), and
\item $ac\equiv bd$ (mod $m$).
\end{enumerate}

Now for your homework problems!
\begin{enumerate}
\item Decide if the following are true or false.
\begin{enumerate}
\item $3\mid 12.$
\item $12\mid 3.$
\item $7\mid -14.$
\item $10\mid 10.$
\item $-1\mid -1.$
\item $-1\mid17.$
\end{enumerate}

\item Explain why $4\mid 0,$ but $0\nmid4.$

\vskip 1in

\item Suppose you know that $a\mid(b+c).$ Can you conclude that $a\mid b$ and $a\mid c$?  Explain your answer.
\vskip 1.5in


\item $a=72,$ and $b=2^m3^n,$ where $m,n>0.$  If $a\mid b,$ what can you conclude about $m$ and $n$?

\vskip 1in

\item Find the following.
\begin{enumerate}
\item 23 (mod 7).
\item $-23$ (mod 7).
\item 12 (mod 12).
\item 1,000,000 (mod 11).
\item $(-3)^{97}$ (mod 4).
\item $(10k-3)$ (mod 5), where $k$ is any integer.
\item 91$p$ (mod 7), where $p$ is any integer.
\end{enumerate}
\item Show, using arithmetic (mod 10), than if $n$ is any positive integer, then $n$ and $n^5$ end in the same digit.
\vskip 2in
\item Find a number $n$ such that $n\equiv2$ (mod 3) and $n\equiv2$ (mod 7).  Can you find {\em every} number $n$ with this property?
\end{enumerate}


\end{document}

