diff --git a/docs/loesungen.pdf b/docs/loesungen.pdf
index 9eb2f73..c21994f 100644
Binary files a/docs/loesungen.pdf and b/docs/loesungen.pdf differ
diff --git a/docs/loesungen.tex b/docs/loesungen.tex
index 8e25c82..64b5b3e 100644
--- a/docs/loesungen.tex
+++ b/docs/loesungen.tex
@@ -53,6 +53,14 @@
 %%            |
 %%            — body/uebung/ueb4.tex;
 %%            |
+%%            — body/uebung/ueb5.tex;
+%%            |
+%%            — body/ska/ska1.tex;
+%%            |
+%%            — body/ska/ska2.tex;
+%%            |
+%%            — body/ska/ska3.tex;
+%%            |
 %%            — body/ska/ska4.tex;
 %%            |
 %%            — body/ska/ska5.tex;
@@ -64,6 +72,8 @@
 %%            — body/quizzes/quiz3.tex;
 %%            |
 %%            — body/quizzes/quiz4.tex;
+%%            |
+%%            — body/quizzes/quiz5.tex;
 %%        |
 %%        — back/index.tex;
 %%            |
@@ -1301,6 +1311,10 @@
 \def\ntrlzero{\mathbb{N}_{0}}
 \def\reellNonNeg{\reell_{+}}
 
+\def\imageinh{\imath}
+\def\ReTeil{\mathop{\mathfrak{R}\text{\upshape e}}}
+\def\ImTeil{\mathop{\mathfrak{I}\text{\upshape m}}}
+
 \def\leer{\emptyset}
 \def\restr#1{\vert_{#1}}
 \def\ohne{\setminus}
@@ -3535,11 +3549,249 @@ für $a,b\in\intgr$.
         Darum gilt $\Phi(n)$ per Induktion für alle $n\in\ntrl$.
     \end{proof}
 
+%% ********************************************************************************
+%% FILE: body/uebung/ueb5.tex
+%% ********************************************************************************
+
+\setcounternach{chapter}{5}
+\chapter[Woche 5]{Woche 5}
+    \label{ueb:5}
+
+\textbf{ACHTUNG.}
+Diese Lösungen dienen \emph{nicht} als Musterlösungen sondern eher als Referenz.
+Hier wird eingehender gearbeitet, als generell verlangt wird.
+Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche vergleichen kann.
+
+%% AUFGABE 5-1
+\let\altsectionname\sectionname
+\def\sectionname{Aufgabe}
+\section[Aufgabe 1]{}
+    \label{ueb:5:ex:1}
+\let\sectionname\altsectionname
+
+Fixiere eine natürliche Zahl $n\in\ntrlzero$.
+Sei $a_{i}\in\{0,1,\ldots,10-1\}$ die eindeutige Zahlen,
+so dass
+
+    \begin{mathe}[mc]{rcl}
+        n &= &\sum_{i\in\ntrlzero}a_{i}10^{i}\\
+    \end{mathe}
+
+gilt.
+
+\begin{enumerate}{\bfseries (a)}
+    %% AUFGABE 5-1a
+    \item
+        \begin{claim*}
+            $3\divides n$ $\Leftrightarrow$ $3\mid\sum_{i\in\ntrlzero}a_{i}$.
+        \end{claim*}
+
+        \begin{proof}
+            Beachte zunächst, dass $10\equiv 1\mod 3$.
+            Also gilt modulo $3$
+
+                \begin{mathe}[mc]{rcccl}
+                    n &\equiv &\sum_{i\in\ntrlzero}a_{i}1^{i} &\equiv &\sum_{i\in\ntrlzero}a_{i}.\\
+                \end{mathe}
+
+            Folglich gilt
+
+                \begin{mathe}[mc]{rcccccl}
+                    3\mid n
+                        &\Longleftrightarrow
+                            &n\equiv 0\mod 3
+                        &\Longleftrightarrow
+                            &\sum_{i\in\ntrlzero}a_{i}\equiv 0\mod 3
+                        &\Longleftrightarrow
+                            &3\mid\sum_{i\in\ntrlzero}a_{i}
+                \end{mathe}
+
+            wie behauptet.
+        \end{proof}
+
+    %% AUFGABE 5-1b
+    \item
+        \begin{claim*}
+            $11\divides n$ $\Leftrightarrow$ $1\mid\sum_{i\in\ntrlzero}(-1)^{i}a_{i}$.
+        \end{claim*}
+
+        \begin{proof}
+            Beachte zunächst, dass $10=-1\mod 11$.
+            Also gilt modulo $11$
+
+                \begin{mathe}[mc]{rcccl}
+                    n &\equiv &\sum_{i\in\ntrlzero}a_{i}(-1)^{i}.\\
+                \end{mathe}
+
+            Folglich gilt
+
+                \begin{mathe}[mc]{rcccccl}
+                    11\mid n
+                        &\Longleftrightarrow
+                            &n\equiv 0\mod 11
+                        &\Longleftrightarrow
+                            &\sum_{i\in\ntrlzero}a_{i}\equiv 0\mod 11
+                        &\Longleftrightarrow
+                            &11\mid\sum_{i\in\ntrlzero}(-1)^{i}a_{i}
+                \end{mathe}
+
+            wie behauptet.
+        \end{proof}
+\end{enumerate}
+
+%% AUFGABE 5-2
+\let\altsectionname\sectionname
+\def\sectionname{Aufgabe}
+\section[Aufgabe 2]{}
+    \label{ueb:5:ex:2}
+\let\sectionname\altsectionname
+
+\begin{enumerate}{\bfseries (a)}
+    %% AUFGABE 5-2a
+    \item
+
+        Seien $a=142$ und $b=84$.
+        Wir berechnen $\ggT(a,b)$ mittels des Euklidischen Algorithmus
+        (siehe \cite[Satz 3.4.7]{sinn2020}).
+
+        \begin{longtable}[mc]{|c|c|}
+            \hline
+            \hline
+                Restberechnung (symbolisch) &Restberechnung (Werte)\\
+            \hline
+            \endhead
+                    $a = b\cdot q_{1} + r_{1}$ &$142 = 84\cdot 1 + 58$\\
+                    $b = r_{1}\cdot q_{2} + r_{2}$ &$84 = 58\cdot 1 + 26$\\
+                    $r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$58 = 26\cdot 2 + 6$\\
+                    $r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$26 = 6\cdot 4 + \boxed{\mathbf{2}}$\\
+                    $r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$6 = 2\cdot 3 + 0$\\
+            \hline
+            \hline
+        \end{longtable}
+
+        Darum gilt $\ggT(a,b)=r_{2}=2$.
+
+    %% AUFGABE 5-2b
+    \item
+
+        \begin{claim}
+            \makelabel{claim:main:ueb:5:ex:2b}
+            Seien $a,b,c\in\intgr$ mit $a,b\neq 0$.
+            Die folgenden Aussagen sind äquivalent:
+
+                \begin{kompaktenum}{\bfseries (i)}[\rtab][\rtab]
+                    \item\punktlabel{1}
+                        $\exists{x,y\in\intgr:~}ax+by=c$
+                    \item\punktlabel{2}
+                        $\ggT(a,b)\divides c$
+                \end{kompaktenum}
+
+        \end{claim}
+
+        \begin{proof}
+            Fixiere zunächst $d:=\ggT(a,b)$.
+            Da $a,b\in\intgr\ohne\{0\}$, ist $d\in\ntrl$ eine wohldefinierte positive Zahl.
+
+            \hinRichtung{1}{2}
+                Angenommen, $ax+by=c$ für ein $x,y\in\intgr$.\\
+                Da $x,y\in\intgr$,
+                erhalten wir $c=ax+by\equiv 0x+0z\equiv 0$ modulo $d$.\\
+                Also $\ggT(a,b)=d\divides c$.
+
+            \hinRichtung{2}{1}
+                Angenommen, $\ggT(a,b)\divides c$.\\
+                Dann existiert ein $k\in\intgr$, so dass $c=k\cdot\ggT(a,b)$.\\
+                Laut des Lemmas von B\'ezout (siehe \cite[Lemma 3.4.8]{sinn2020})
+                existiere nun $u,v\in\intgr$, so dass $\ggT(a,b)=au+bv$.\\
+                Daraus folgt ${c=k\cdot\ggT(a,b)=aku+bkv}$.\\
+                Da $ku,kv\in\intgr$, haben wir \punktcref{1} bewiesen.
+        \end{proof}
+\end{enumerate}
+
+%% AUFGABE 5-3
+\let\altsectionname\sectionname
+\def\sectionname{Aufgabe}
+\section[Aufgabe 3]{}
+    \label{ueb:5:ex:3}
+\let\sectionname\altsectionname
+
+\begin{enumerate}{\bfseries (a)}
+    %% AUFGABE 5-3a
+    \item
+
+        Sei $H:=\intgr/2\intgr$ die (abelsche) Gruppe von Restklassen modulo $2$ unter Addition.\\
+        Sei $G:=H\times H$ mit Neutralelement $e=([0],[0])$ und versehen mit der Produktstruktur.\\
+        Als Produkt von (abelschen) Gruppen ist $G$ automatisch eine (abelsche) Gruppe.
+        Und offensichtlich hat $G$ genau $|G|=|H\times H|=|H|\cdot|H|=2\cdot 2=4$ Elemente.\\
+        Es bleibt \textbf{zu zeigen}, dass $\forall{a\in G:~}a\ast a=e$.\\
+        Sei also $a=([k],[j])\in H\times H=G$ ein beliebiges Element.
+        Es gilt
+
+            \begin{mathe}[mc]{rcl}
+                a\ast a
+                    &= &([k],[j])\ast([k],[j])\\
+                    &= &([k]+[k],[j]+[j])\\
+                    &= &([k+k],[j+j])\\
+                    &= &([2k],[2j])
+                    =([0],[0])
+                    =e,
+            \end{mathe}
+
+        da $2\equiv 0\mod 2$.\\
+        Also ist unsere Konstruktion von $G$ ein passendes Beispiel.
+
+    %% AUFGABE 5-3b
+    \item
+        \begin{claim*}
+            Sei $(G,\ast,e)$ eine Gruppe.
+            Angenommen, $\forall{a\in G:~}a\ast a=e$.
+            Dann ist $G$ kommutativ.
+        \end{claim*}
+
+        \begin{proof}
+            Beachte zunächst, dass wegen Eindeutigkeit des Inverses
+            die Annahme zu
+
+                \begin{mathe}[mc]{c}
+                    \eqtag[eq:1:ueb:5:ex:3]
+                    \forall{a\in G:~}a^{-1}=a
+                \end{mathe}
+
+            äquivalent ist.\\
+            \textbf{Zu zeigen:} Für alle $a,b\in G$ gilt $a\ast b=b\ast a$.\\
+            Seien also $a,b\in G$ beliebige Elemente.
+            Es gilt
+
+                \begin{mathe}[mc]{rcccccl}
+                    a\ast b
+                        &\eqcrefoverset{eq:1:ueb:5:ex:3}{=}
+                            &a^{-1}\ast b^{-1}
+                        &= &(b\ast a)^{-1}
+                        &\eqcrefoverset{eq:1:ueb:5:ex:3}{=}
+                            &b\ast a.\\
+                \end{mathe}
+
+            Also ist $G$ eine kommutative Gruppe.
+        \end{proof}
+\end{enumerate}
+
 \setcounternach{part}{2}
 \part{Selbstkontrollenaufgaben}
 
     \def\chaptername{SKA Blatt}
 
+%% ********************************************************************************
+%% FILE: body/ska/ska1.tex
+%% ********************************************************************************
+
+%% ********************************************************************************
+%% FILE: body/ska/ska2.tex
+%% ********************************************************************************
+
+%% ********************************************************************************
+%% FILE: body/ska/ska3.tex
+%% ********************************************************************************
+
 %% ********************************************************************************
 %% FILE: body/ska/ska4.tex
 %% ********************************************************************************
@@ -5040,14 +5292,14 @@ Um diese Urteil also leichter treffen zu können ersetzen wir die Elemente durch
                 \pgfmathsetmacro\vabst{1}
                 \pgfmathsetmacro\rad{8mm}
 
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-                \node[rectangle, label=left:{$(1\ 2)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:white,255}, draw] (g_1) at (0*\habst, -2*\vabst) {};
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-                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:white,64}, draw] (gh_0_0) at (1*\habst, -1*\vabst) {};
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-                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:white,255}, draw] (gh_1_0) at (1*\habst, -2*\vabst) {};
-                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:white,64}, draw] (gh_1_1) at (2*\habst, -2*\vabst) {};
+                \node[rectangle, label=left:{$e=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (g_0) at (0*\habst, -1*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (h_0) at (1*\habst, 0*\vabst) {};
+                \node[rectangle, label=left:{$(1\ 2)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (g_1) at (0*\habst, -2*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (h_1) at (2*\habst, 0*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_0_0) at (1*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_0_1) at (2*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_1_0) at (1*\habst, -2*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_1_1) at (2*\habst, -2*\vabst) {};
             \end{tikzpicture}
         }
         \hraum
@@ -5057,54 +5309,54 @@ Um diese Urteil also leichter treffen zu können ersetzen wir die Elemente durch
                 \pgfmathsetmacro\vabst{1}
                 \pgfmathsetmacro\rad{8mm}
 
-                \node[rectangle, label=left:{$e=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:white,64}, draw] (g_0) at (0*\habst, -1*\vabst) {};
-                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:white,64}, draw] (h_0) at (1*\habst, 0*\vabst) {};
-                \node[rectangle, label=left:{$(2\ 3)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:white,102}, draw] (g_1) at (0*\habst, -2*\vabst) {};
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-                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:white,64}, draw] (gh_5_5) at (6*\habst, -6*\vabst) {};
+                \node[rectangle, label=left:{$e=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (g_0) at (0*\habst, -1*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (h_0) at (1*\habst, 0*\vabst) {};
+                \node[rectangle, label=left:{$(2\ 3)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (g_1) at (0*\habst, -2*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (h_1) at (2*\habst, 0*\vabst) {};
+                \node[rectangle, label=left:{$(1\ 2)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (g_2) at (0*\habst, -3*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (h_2) at (3*\habst, 0*\vabst) {};
+                \node[rectangle, label=left:{$(1\ 2\ 3)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (g_3) at (0*\habst, -4*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (h_3) at (4*\habst, 0*\vabst) {};
+                \node[rectangle, label=left:{$(1\ 3\ 2)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (g_4) at (0*\habst, -5*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (h_4) at (5*\habst, 0*\vabst) {};
+                \node[rectangle, label=left:{$(1\ 3)=:$}, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (g_5) at (0*\habst, -6*\vabst) {};
+                \node[rectangle, line width=2pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (h_5) at (6*\habst, 0*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_0_0) at (1*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_0_1) at (2*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_0_2) at (3*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_0_3) at (4*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_0_4) at (5*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_0_5) at (6*\habst, -1*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_1_0) at (1*\habst, -2*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_1_1) at (2*\habst, -2*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_1_2) at (3*\habst, -2*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_1_3) at (4*\habst, -2*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_1_4) at (5*\habst, -2*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_1_5) at (6*\habst, -2*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_2_0) at (1*\habst, -3*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_2_1) at (2*\habst, -3*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_2_2) at (3*\habst, -3*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_2_3) at (4*\habst, -3*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_2_4) at (5*\habst, -3*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_2_5) at (6*\habst, -3*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_3_0) at (1*\habst, -4*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_3_1) at (2*\habst, -4*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_3_2) at (3*\habst, -4*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_3_3) at (4*\habst, -4*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_3_4) at (5*\habst, -4*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_3_5) at (6*\habst, -4*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_4_0) at (1*\habst, -5*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_4_1) at (2*\habst, -5*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_4_2) at (3*\habst, -5*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_4_3) at (4*\habst, -5*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_4_4) at (5*\habst, -5*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_4_5) at (6*\habst, -5*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_5_0) at (1*\habst, -6*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,200}, draw] (gh_5_1) at (2*\habst, -6*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_5_2) at (3*\habst, -6*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_5_3) at (4*\habst, -6*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_5_4) at (5*\habst, -6*\vabst) {};
+                \node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_5_5) at (6*\habst, -6*\vabst) {};
             \end{tikzpicture}
         }
         \hraum
@@ -5386,6 +5638,47 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
         \end{proof}
 \end{enumerate}
 
+%% ********************************************************************************
+%% FILE: body/quizzes/quiz5.tex
+%% ********************************************************************************
+
+\setcounternach{chapter}{5}
+\chapter[Woche 5]{Woche 5}
+    \label{quiz:5}
+
+\begin{claim*}
+    Seien $n\in\ntrlpos$ und $p\in\mathbb{P}$ mit $n<p\leq 2n$.
+    Dann gilt $p\divides\begin{svector}2n\\n\\\end{svector}$.
+\end{claim*}
+
+\begin{proof}
+    Aus $n<p\leq 2n$, d.\,h. $p\in\{n+1,n+2,\ldots,2n\}$, folgt (i)~$p\divides\prod_{i=n+1}^{2n}i$.\\
+    Es gilt nun
+
+        \begin{mathe}[mc]{rcccccl}
+            \eqtag[eq:1:quiz:5:ex:1]
+                \prod_{i=n+1}^{2n}i
+                &= &\dfrac{\prod_{i=1}^{2n}i}{n!}
+                &= &n!\dfrac{(2n)!}{n!(2n-n)!}
+                &= &n!\begin{vector}2n\\n\\\end{vector}.\\
+        \end{mathe}
+
+    Aus (i) und \eqcref{eq:1:quiz:5:ex:1} folgt also
+    (ii)~$p\divides \begin{svector}2n\\n\\\end{svector}\cdot n!$.\\
+    Beachte, dass $p$ eine Primzahl ist und ${n!,\begin{svector}2n\\n\\\end{svector}\in\intgr}$.\\
+    Aus (ii) und \cite[Satz 3.4.14]{sinn2020} folgt also
+        $p\divides\begin{svector}2n\\n\\\end{svector}$ oder $p\divides n!$.\\
+
+    \fbox{Angenommen, $p\ndivides\begin{svector}2n\\n\\\end{svector}$.}\\
+    Dann muss laut des o.\,s. Arguments ${p\divides n!(=\prod_{i=1}^{n}i)}$ gelten.\\
+    Eine weitere Anwendung von \cite[Satz 3.4.14]{sinn2020} liefert,
+    dass ${p\divides i_{0}}$ für ein $i_{0}\in\{1,2,\ldots,n\}$.\\
+    Aber dann gilt $1\leq p\leq i_{0}\leq n$.
+    Das widerspricht der Voraussetzung, dass $n<p$.\\
+    Darum stimmt die Annahme nicht.
+    Das heißt, $p\divides\begin{svector}2n\\n\\\end{svector}$.
+\end{proof}
+
 %% ********************************************************************************
 %% FILE: back/index.tex
 %% ********************************************************************************