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{\Large METU MATH 365, EXAM 1}\hspace{\stretch{1}} 
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Monday, November 8, 2010, at 13:40

Instructor: David Pierce

%\vspace{1ex}

%\noindent{\large first and last NAME:}

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%\noindent{\large Student number:}

\vspace{2ex}

\parbox{13cm}{\emph{Instructions:} 
Write your full name, and your student number, at the top of this page.
Please work carefully.  There are 6 numbered problems on 4 pages.
It should be
obvious to the grader how to read your solutions.  }

\begin{problem}
Let $\vnn=\{0,1,2,\dots\}$.  All variables in this problem range over $\vnn$.  Given $a$ and $b$ such that $a\neq0$, we define
\begin{equation*}
\rem{b,a}=r,
\end{equation*}
if $b=ax+r$ for some $x$, and $r<a$.
\begin{enumerate}
\item
Prove $\rem{a+b,n}=\rem{\rem{a,n}+\rem{b,n},n}$.
\item
Prove $\rem{ab,n}=\rem{\rem{a,n}\cdot\rem{b,n},n}$.
\end{enumerate}
\end{problem}

\newpage

\begin{problem}
Find integers $k$ and $\ell$, both greater than $1$, such that, for
all positive integers~$n$, 
\begin{equation*}
k\divides 1965^{10n}+\ell.
\end{equation*}
\end{problem}
\vfill
\begin{problem}
Find two positive integers $a$ and $b$ such that, for all integers $m$ and $n$, the integer $am-bn$ is a solution of the congruences
\begin{align*}
x&\equiv m\pmod{999},&
x\equiv n\pmod{1001}.
\end{align*}
\end{problem}
\vfill
\newpage
\begin{problem}
Letting  $n=\sum_{j=1}^{408}j$, find an integer $k$ such that $0\leq k<409$ and
\begin{equation*}
408!\equiv k\pmod n.
\end{equation*}
\end{problem}
\vfill
\begin{problem}
With justification, find an integer $n$, greater than $1$, such that,
for all integers $a$, 
\begin{equation*}
a^n\equiv a\pmod{1155}.
\end{equation*}
\end{problem}
\vfill
\newpage
\begin{problem}
Let $\N=\{1,2,3,\dots\}$.
Suppose all we know about this set is:
\begin{compactenum}[(i)]
\item
proofs by induction are possible;
\item
addition can be defined on $\N$, and it satisfies
\begin{align*}
x+y&=y+x,&x+(y+z)&=(x+y)+z;
\end{align*}
\item
multiplication can be defined by
\begin{align*}
x\cdot 1&=x,&x\cdot(y+1)&=x\cdot y+x.
\end{align*}
\end{compactenum}
Prove
\begin{equation*}
x\cdot y=y\cdot x.
\end{equation*}
 
\end{problem}

\end{document}
