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\usepackage{hfoldsty,paralist,multicol}
\usepackage[turkish]{babel}
\usepackage{amsmath,url,amsthm,amssymb}
\newcommand{\divides}{\mid}
\newcommand{\ndivides}{\nmid}
\DeclareMathOperator{\ebob}{ebob}
\renewcommand{\gcd}{\ebob}

\usepackage{nicefrac}
\newcommand{\matfrac}[2]{\nicefrac{#1}{#2}}

\theoremstyle{definition}
\newtheorem{problem}{Problem}
\theoremstyle{remark}
\newtheorem*{solution}{\c C\"oz\"um}


\usepackage{verbatim}
%\let\solution=\comment
%\let\endsolution=\endcomment

\pagestyle{empty}
\begin{document}
%\subtitle{S\i nav I}
\title{Say\i lar Kuram\i na Giri\c s}
\subtitle{{\scshape MAT} \newstylenums{111}, S\i nav I \itshape \c c\"oz\"umleri}
\author{David Pierce}
\publishers{Matematik B\"ol\"um\"u, MSGS\"U}
\date{24 Ekim 2017}
%\publishers{Matematik B\"ol\"um\"u, MSGS\"U}
%\mbox{}\\
%\url{dpierce@msgsu.edu.tr}\\
%\url{http://mat.msgsu.edu.tr/~dpierce/}

\maketitle\thispagestyle{empty}

Bu s\i navda her say\i\ tamsay\i d\i r.
  \c C\"oz\"um y\"ontemlerinizi d\"u\c s\"unerek se\c cin,
  ve \c c\"oz\"umlerinizi net bir \c sekilde yaz\i n.
  M\"umk\"unse cevaplar\i n\i z\i\ kontrol edin.
	
	Her s\i navdaki gibi,
	ba\c ska kimseyle konu\c smay\i n;
	ba\c ska kimsenin ka\u g\i d\i n\i\ g\"ormeye \c cal\i\c smay\i n;
	kendi ka\u g\i d\i n\i z\i\ ba\c ska kimseye g\"ostermeyin,
	getirdi\u giniz notlar\i, kitaplar\i, veya cihazlar\i\ kullanmay\i n.
  
	\.Iyi \c cal\i\c smalar dilerim.

\begin{problem}
Tan\i ma g\"ore $a\divides b$ 
ancak ve ancak $ax=b$ denklemi \c c\"oz\"ulebilir.
A\c sa\u g\i daki gerektirmelerin her birini ya kan\i tlay\i n
ya da kar\c s\i t \"ornek ile \c c\"ur\"ut\"un.
\begin{multicols}{2}
\begin{enumerate}[(a)]
\item
$ab\divides c$ ise $a\divides c$ ve $b\divides c$.
\item
$a\divides bc$ ise $a\divides b$ veya $a\divides c$.
\end{enumerate}
\end{multicols}
\end{problem}

\begin{solution}
\begin{enumerate}[(a)]
\item
$ab\divides c$ olsun.
O zaman bir $x$ i\c cin $abx=c$.
\"Ozellikle $c=a(bx)=b(ax)$,
dolay\i s\i yla $a\divides c$ ve $b\divides c$.
\item
$4\divides 2\cdot2$ ama $4\ndivides 2$.
\end{enumerate}
\end{solution}

\newpage

\begin{problem}
\"Ozyineli tan\i ma g\"ore $t_1=1$
ve her $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
t_{n+1}=t_n+n+1.	
\end{equation*}
Kan\i tlad\i\u g\i m\i z teoreme g\"ore
\begin{equation*}
t_n=\textstyle\frac12n\cdot(n+1).
\end{equation*}
E\u ger $t_0=0$ tan\i mlan\i rsa
ayn\i\ e\c sitlikler $n=0$ durumunda da do\u grudur.
\c Simdi
\begin{align*}
	p_1&=1,&p_{n+1}&=p_n+3n+1
\end{align*}
olsun.
Her $n$ sayma say\i s\i\ i\c cin
a\c sa\u g\i daki e\c sitlikleri kan\i tlay\i n.
\begin{multicols}{2}
    \begin{enumerate}[(a)]
  \item
    $p_n=\frac32n^2-\frac12n$.
  \item
    $p_n=n^2+t_{n-1}$.
  \end{enumerate}
\end{multicols}
\end{problem}

\begin{solution}
\begin{enumerate}[(a)]
\item
$n=1$ ise
$\frac32n^2-\frac12n=\frac32-\frac12=1=p_1=p_n$.
\.Iddia $n=m$ durumunda do\u gru ise
\begin{align*}
p_{m+1}
&=p_m+3m+1&&\text{[tan\i m]}\\
&=\textstyle\frac32m^2-\textstyle\frac12m+3m+1&&\text{[hipotez]}\\
%&=\frac32(m^2+2m+1)-\frac12m-\frac12&&\\
&=\textstyle\frac32(m^2+2m+1)-\textstyle\frac12(m+1)&&\\
&=\textstyle\frac32(m+1)^2-\textstyle\frac12(m+1).
\end{align*}
T\"umevar\i m ile istedi\u gimiz sonu\c c \c c\i kar.
\item
$\begin{aligned}[t]
n^2+t_{n-1}
&=n^2+\textstyle\frac12(n-1)n&&\text{[teorem]}\\
&=n^2+\textstyle\frac12n^2-\textstyle\frac12n&&\text{[hesaplama]}\\
&=\textstyle\frac32n^2-\textstyle\frac12n&&\text{[hesaplama]}\\
&=p_n.&&\text{[(a) par\c cas\i]}
\end{aligned}$
\end{enumerate}
\end{solution}

\newpage
\begin{problem}
Tamsay\i larda
\begin{align*}
	\gcd(a,b)&=\gcd(b,a),&
	\gcd(a,b)&=\gcd(a\pm kb,b)
\end{align*}
e\c sitliklerini biliyoruz.
A\c sa\u g\i daki e\c sitliklerin her birini ya kan\i tlay\i n
ya da kar\c s\i t \"ornek ile \c c\"ur\"ut\"un.
\begin{multicols}{2}
\begin{enumerate}[(a)]
\item
$\gcd(a,bc)=b\cdot\gcd(a,c)$.
\item
$\gcd(2a+b,a+b)=\gcd(a,b)$.
\end{enumerate}
\end{multicols}
\end{problem}

\begin{solution}
\begin{enumerate}[(a)]
\item
$\gcd(1,2\cdot1)=1$ ama $2\cdot\gcd(1,1)=2$.
\item
  $2a+b-(a+b)=a$ oldu\u gundan
  \begin{equation*}
 \gcd(2a+b,a+b)=\gcd(a,a+b)=\gcd(a,b).   
  \end{equation*}
\end{enumerate}

\end{solution}

\newpage

\begin{problem}
  A\c sa\u g\i dakileri bulun.
\begin{enumerate}[(a)]
\item
$\gcd(942,180)$.
\item
$\gcd(942,180)=942x+180y$ denkleminin bir \c c\"oz\"um\"u.
\item
$\cfrac{942}{180}=a+\cfrac1{b+\cfrac1{c+\cfrac1d}}$
e\c sitli\u gini sa\u glayan
$a$, $b$, $c$, ve $d$.
\end{enumerate}
\end{problem}

\begin{solution}
\begin{enumerate}[(a)]
\item
$\left\{\begin{gathered}
942=180\cdot5+42,\\
180= 42\cdot4+12,\\
 42= 12\cdot3+ 6,\\
 12=  6\cdot2,
\end{gathered}\right\}$
dolay\i s\i yla \fbox{$\gcd(942,180)=6$.}
\item
$\left\{\begin{aligned}
6
&=42-12\cdot3\\
&=42-(180-42\cdot4)\cdot3\\
&=42\cdot13-180\cdot3\\
&=(942-180\cdot5)\cdot13-180\cdot3\\
&=942\cdot13-180\cdot68
\end{aligned}\right\}$
ve b\"oylece $(x,y)$, \fbox{$(13,-68)$} olabilir.
Kontrol:
$\begin{array}[t]{r*4{@{}c}}
      & &9&4&2\\
\times& & &1&3\\\hline
      &2&8&2&6\\
      &9&4&2& \\\hline
     1&2&2&4&6
\end{array}\quad
\begin{array}[t]{r*4{@{}c}}
      & &1&8&0\\
\times& & &6&8\\\hline
      &1&4&4&0\\
     1&0&8&0& \\\hline
     1&2&2&4&0
\end{array}$
\item
$\cfrac{942}{180}=5+\cfrac1{4+\cfrac1{3+\cfrac12}}$
\end{enumerate}
\end{solution}

\end{document}
