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\newtheorem*{remark}{Not}

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\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 II  \itshape \c c\"oz\"umleri}
\author{David Pierce}
\publishers{Matematik B\"ol\"um\"u, MSGS\"U}
\date{12 Aral\i k 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}

  \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 \fbox{cevaplar\i n\i z\i\ kontrol edin.}
Yazd\i\u g\i n\i z \fbox{her c\"umlenin, okura a\c c\i k olan nedeni olmal\i.}

%Her say\i\ tamsay\i d\i r.
	
	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}
A\c sa\u g\i daki denklik,
bir $\sim$ ba\u g\i nt\i s\i n\i n tan\i m\i\ olsun:
\begin{equation*}
    a\sim b\iff a+b\text{ \c cifttir}.
\end{equation*}
Bu $\sim$ ba\u g\i nt\i s\i n\i n 
bir denklik ba\u g\i nt\i s\i\ oldu\u gunu g\"osterin.
\end{problem}

        \begin{solution}\sloppy
Verilen ba\u g\i nt\i n\i n
\begin{inparaenum}[(i)]
\item
yans\i mal\i,
\item
simetrik ve 
\item
ge\c ci\c sli oldu\u gunu g\"osterece\u giz.
\end{inparaenum}
          \begin{enumerate}[(i)]
            \item
          $a+a=2a$ ve $2a$ \c cifttir,
dolay\i s\i yla $a\sim a$.
\item
$b+a=a+b$ dolay\i s\i yla
$a+b$ \c cift ise $b+a$ da \c cifttir.
B\"oylece $a\sim b$ ise $b\sim a$.
\item
$a+b$ ve $b+c$ \c cift ise
$a+c$ da \c cifttir \c c\"unk\"u
\begin{equation*}
  a+c=(a+b)+(b+c)-2b.
\end{equation*}
Sonu\c c olarak $a\sim b$ ve $b\sim c$ ise $a\sim c$.
\end{enumerate}
\end{solution}

\begin{remark}
Ba\c ska \c c\"oz\"umler de m\"umk\"und\"ur.
\"Orne\u gin tan\i ma g\"ore
\begin{align*}
  a\sim b
&\iff2\divides a+b\\
&\iff a\equiv-b\pmod 2\\
&\iff a\equiv b\pmod 2
\end{align*}
\c c\"unk\"u $2$'ye g\"ore $-1\equiv1$.
Derste g\"osterdi\u gimiz gibi,
sabit bir mod\"ule g\"ore kalanda\c sl\i k,
bir denklik ba\u g\i nt\i s\i d\i r.
\end{remark}

%\newpage

\begin{problem}\mbox{}
  \begin{enumerate}[(a)]
  \item 
  $a\divides b$ ve $b\divides c$ ise $a\divides c$ g\"osterin.
\item
Her $n$ sayma say\i s\i\ i\c cin
$\sum_{k=1}^nk!\cdot k=(n+1)!-1$ g\"osterin.
\end{enumerate}
\end{problem}

\begin{solution}
  \begin{asparaenum}[(a)]
    \item
  Varsay\i ma g\"ore baz\i\ $x$ ve $y$ i\c cin
  \begin{align*}
    ax&=b,& by&=c.
  \end{align*}
Sonu\c c olarak $c=(ax)y=a(xy)$,
dolay\i s\i yla $a\divides c$.
\item
  T\"umevar\i m kullanaca\u g\i z.
  \.Iki ad\i m vard\i r.
  \begin{compactenum}[(i)]
    \item
  $n=1$ ise iddia do\u grudur, \c c\"unk\"u
  \begin{equation*}
    \sum_{k=1}^1k!\cdot k=1!\cdot1=1
  \end{equation*}
  ve ayn\i\ zamanda $(1+1)!-1=2!-1=2-1=1$.
\item
  \c Simdi $n=m$ durumunda iddia do\u gru varsay\i l\i rsa,
o zaman
\begin{align*}
   \sum_{k=1}^{m+1}k!\cdot k
&=\sum_{k=1}^mk!\cdot k+(m+1)!\cdot(m+1)&&\text{[tan\i m]}\\
&=(m+1)!-1+(m+1)!\cdot(m+1)            &&\text{[hipotez]}\\
&=(m+1)!\cdot(m+2)-1                   &&\text{[hesap]}\\
&=(m+2)!-1                             &&\text{[tan\i m]},
\end{align*}
dolay\i s\i yla iddia $n=m+1$ durumunda do\u grudur.
  \end{compactenum}
  T\"umevar\i mdan iddia her durumda do\u grudur.
\end{asparaenum}
\end{solution}

\begin{remark}
  Birinci kan\i tta $a\divides b$ ifadesinin anlam\i,
  \emph{baz\i} $x$ i\c cin $ax=b$.
  \.Ikinci kan\i tta
  e\c sitlikler \emph{s\i ras\i yla} anla\c s\i labilmeli.
\end{remark}

%\newpage
        \begin{problem}
$997$ asald\i r.
          \begin{enumerate}[(a)]
            \item
          $\mathbb Z_{997}$'de $100$'\"un tersini bulun, yani
          \begin{equation*}
            100x\equiv1\pmod{997}
          \end{equation*}
kalanda\c sl\i\u g\i n\i\ \c c\"oz\"un.
\item
$\mathbb Z_{997}$'de $100^{995}$ kuvvetini hesaplay\i n, yani
\begin{equation*}
  100^{995}\equiv x\pmod{997}\And 0<x<997
\end{equation*}
sistemini \c c\"oz\"un.
\end{enumerate}
\end{problem}

\begin{solution}
  \begin{asparaenum}[(a)]
  \item
    \"Oklid Algoritmas\i n\i\ kullanaca\u g\i z.
    \begin{gather*}
      997=100\cdot9+97,\\
100=97\cdot1+3,\\
97=3\cdot32+1,
    \end{gather*}
dolay\i s\i yla
    \begin{align*}
      1
&=97-3\cdot32=97-(100-97)\cdot32\\
&=97\cdot33-100\cdot32=(997-100\cdot9)\cdot33-100\cdot32\\
&=997\cdot33-100\cdot329.
    \end{align*}
    Kontrol edelim:
    $997\cdot33=(1000-3)\cdot33$
    \begin{equation*}
  =33000-99=32901
	=1+329\cdot100.      
    \end{equation*}
B\"oylece
$1\equiv100\cdot(-329)\equiv100\cdot668\pmod{997}$,
yani
\begin{equation*}
  100^{-1}\equiv668\pmod{997}.
\end{equation*}
\item
Fermat Teoreminden
\begin{equation*}
1\equiv100^{996}\equiv100^{995}\cdot100\pmod{997}
\end{equation*}
dolay\i s\i yla (a) \c s\i kk\i ndan
$100^{995}\equiv668\pmod{997}$.
  \end{asparaenum}
\end{solution}
%\newpage

\begin{problem}
Ya a\c sa\u g\i daki \"onermeyi kan\i tlay\i n, 
ya da kar\c s\i t bir \"ornek ile 
\"onermenin yanl\i\c s oldu\u gunu kan\i tlay\i n.
  \begin{quote}\itshape
    Bir $p$ tek asal say\i s\i\ i\c cin $\frac12(p-1)=n$ olsun.
Her $a$ tamsay\i s\i\ i\c cin, e\u ger $p$'ye g\"ore
$a^n\equiv-1$
ise, o zaman $a$, $p$'nin ilkel bir k\"ok\"ud\"ur.
  \end{quote}
\end{problem}

\begin{solution}
  $7$ tek ve asald\i r,
$\frac12(7-1)=3$ ve 
\begin{equation*}
6^3\equiv(-1)^3\equiv-1\pmod 7.
\end{equation*}
Ayn\i\ zamanda
$6^2\equiv(-1)^2\equiv1\pmod 7$,
ve $2<7-1$,
dolay\i s\i yla $6$,
$7$'nin ilkel bir k\"ok\"u de\u gildir.
B\"oylece \"onerme yanl\i\c st\i r.
\end{solution}

\end{document}
