% LaTeX solution template for Math 235
% -----------------------------------------------------------------

% Like most advanced LaTeX files, this one begins with a lot of
% boilerplate. You don't need to understand (or even read) most of it.
% All you need to do is fill in your name, email address,
% and the number of the pset. (Search for "METADATA" to find the place
% for this.) Then, you can go straight to the "EXERCISE 1"
% section and start writing your solutions.
% The "VARIOUS USEFUL COMMANDS" section is probably worth taking a
% look at at some point.

% If you are here to learn about LaTeX, search for
% "START READING HERE" and start reading there.

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%	PACKAGES AND OTHER DOCUMENT CONFIGURATIONS
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\documentclass[paper=a4, fontsize=12pt]{scrartcl} % A4 paper and 12pt font size
\usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs
\usepackage[english]{babel} % English language/hyphenation
\usepackage{amsmath,amsfonts,amsthm,amssymb} % Math packages
\usepackage{mathrsfs}    % More math packages
\usepackage{sectsty}  % Allows customizing section commands
\allsectionsfont{\centering \normalfont\scshape} % Make all section titles centered, the default font and small caps %remove this to left align section tites
\usepackage{hyperref} % Turns cross-references into hyperlinks,
                      % and defines \url and \href commands.
\usepackage{graphicx} % For embedding graphics files.
\usepackage{framed}   % For the "leftbar" environment used below.
\usepackage{ifthen}   % Used for the \powset command below.
\usepackage{lastpage} % for counting the number of pages
\usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage}
\usepackage[height=10in,a4paper,hmargin={1in,0.8in}]{geometry}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{tikz}     % This is a powerful tool to draw vector
                      % graphics inside LaTeX. In particular, you can
                      % use it to draw graphs.
\usepackage{verbatim} % For the "verbatim" environment, in which
                      % special symbols can be used freely without
                      % confusing the compiler. (And it's typeset in
                      % a constant-width font.)
                      % Useful, e.g., for quoting code (or ASCII art).

%\numberwithin{table}{section} % Number tables within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)

\setlength\parindent{20pt} % Makes indentation for paragraphs longer.
                           % This makes paragraphs stand out more.

%----------------------------------------------------------------------------------------
%	VARIOUS USEFUL COMMANDS
%----------------------------------------------------------------------------------------
% The commands below might be convenient. For example, you probably
% prefer to write $\powset[2]{V}$ for the set of $2$-element subsets
% of $V$, rather than writing $\mathcal{P}_2(V)$.
% Notice that you can easily define your own commands like this.
% Caveat: Some of these commands need to be properly "guarded" when
% they occur in subscripts or superscripts. So you should not write
% $K_\CC$, but rather $K_{\CC}$.
\newcommand{\CC}{\mathbb{C}} % complex numbers
\newcommand{\RR}{\mathbb{R}} % real numbers
\newcommand{\QQ}{\mathbb{Q}} % rational numbers
\newcommand{\NN}{\mathbb{N}} % nonnegative integers
\newcommand{\Z}[1]{\mathbb{Z}/#1\mathbb{Z}} % integers modulo k
                                            % (syntax: "\Z{k}")
\newcommand{\ZZ}{\mathbb{Z}} % integers
\newcommand{\id}{\operatorname{id}} % identity map
\newcommand{\lcm}{\operatorname{lcm}}
% Lowest common multiple. For historical reasons, LaTeX has a \gcd
% command built in, but not an \lcm command. The preceding line
% rectifies that.
\newcommand{\rev}{\operatorname{rev}} % reversal of a walk
\newcommand{\powset}[2][]{\ifthenelse{\equal{#2}{}}{\mathcal{P}\left(#1\right)}{\mathcal{P}_{#1}\left(#2\right)}}
% $\powset[k]{S}$ stands for the set of all $k$-element subsets of
% $S$. The argument $k$ is optional, and if not provided, the result
% is the whole powerset of $S$.
\newcommand{\set}[1]{\left\{ #1 \right\}}
% $\set{...}$ compiles to {...} (set-brackets).
\newcommand{\abs}[1]{\left| #1 \right|}
% $\abs{...}$ compiles to |...| (absolute value, or size of a set).
\newcommand{\tup}[1]{\left( #1 \right)}
% $\tup{...}$ compiles to (...) (parentheses, or tuple-brackets).
\newcommand{\ive}[1]{\left[ #1 \right]}
% $\ive{...}$ compiles to [...] (Iverson bracket, aka truth value).
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
% $\floor{...}$ compiles to |_..._| (floor function).
\newcommand{\verts}[1]{\operatorname{V}\left( #1 \right)}
% $\verts{...}$ compiles to V(...) (vertex set of a graph/digraph).
\newcommand{\edges}[1]{\operatorname{E}\left( #1 \right)}
% $\edges{...}$ compiles to E(...) (edge set of a graph).
\newcommand{\arcs}[1]{\operatorname{A}\left( #1 \right)}
% $\arcs{...}$ compiles to A(...) (arc set of a digraph).
\newcommand{\lf}[2]{#1^{\underline{#2}}}
% $\lf{...1}{...2}$ compiles to $...1^{\underline{...2}}$.
% This is a notation for the falling factorial.
\newcommand{\underbrack}[2]{\underbrace{#1}_{\substack{#2}}}
% $\underbrack{...1}{...2}$ yields
% $\underbrace{...1}_{\substack{...2}}$. This is useful for doing
% local rewriting transformations on mathematical expressions with
% justifications. For example, try this out:
% $ \underbrack{(a+b)^2}{= a^2 + 2ab + b^2 \\ \text{(by the binomial formula)}} $
\newcommand{\horrule}[1]{\rule{\linewidth}{#1}} % Create horizontal rule command with 1 argument of height
\newcommand{\nnn}{\nonumber\\} % Don't number this line in an "align" environment, and move on to the next line.
\newcommand{\explain}[1]{\tup{\hspace{-0.5pc}\text{\begin{tabular}{c}#1\end{tabular}\hspace{-0.5pc}}}}
% $\explain{...}$ interprets the "..." as text (you can
% nest some math inside by putting it in $...$'s as you
% would in any other text block) and puts it in parentheses.

%----------------------------------------------------------------------------------------
%	MAKING SUMMATION SIGNS ALWAYS PUT THEIR BOUNDS ABOVE AND BELOW
%	THE SIGN
%----------------------------------------------------------------------------------------
% The following are hacks to ensure that sums (such as
% $\sum_{k=1}^n k$) always put their bounds (i.e., the $k=1$ and the
% $n$) underneath and above the sign, as opposed to on its right.
% Same for products (\prod), set unions (\bigcup) and set
% intersections (\bigcap). Remove the 8 lines below if you do not want
% this behavior.
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\cupnonlimits\bigcup
\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}

%----------------------------------------------------------------------------------------
%	ENVIRONMENTS
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% The incantations below define how theorem environments
% (\begin{theorem} ... \end{theorem}) and their likes will look like.
\newtheoremstyle{plainsl}% <name>
  {8pt plus 2pt minus 4pt}% <Space above>
  {8pt plus 2pt minus 4pt}% <Space below>
  {\slshape}% <Body font>
  {0pt}% <Indent amount>
  {\bfseries}% <Theorem head font>
  {.}% <Punctuation after theorem head>
  {5pt plus 1pt minus 1pt}% <Space after theorem headi>
  {}% <Theorem head spec (can be left empty, meaning `normal')>

% Environments which make the text inside them slanted:
\theoremstyle{plainsl}
  \newtheorem{theorem}{Theorem}[section]
  \newtheorem{proposition}[theorem]{Proposition}
  \newtheorem{lemma}[theorem]{Lemma}
  \newtheorem{corollary}[theorem]{Corollary}
  \newtheorem{conjecture}[theorem]{Conjecture}
% Environments that don't:
\theoremstyle{definition}
  \newtheorem{definition}[theorem]{Definition}
  \newtheorem{example}[theorem]{Example}
  \newtheorem{exercise}[theorem]{Exercise}
  \newtheorem{examples}[theorem]{Examples}
  \newtheorem{algorithm}[theorem]{Algorithm}
  \newtheorem{question}[theorem]{Question}
 \theoremstyle{remark}
  \newtheorem{remark}[theorem]{Remark}
\newenvironment{statement}{\begin{quote}}{\end{quote}}
\newenvironment{fineprint}{\begin{small}}{\end{small}}

%----------------------------------------------------------------------------------------
%	METADATA
%----------------------------------------------------------------------------------------
\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME HERE
\newcommand{\mymail}{darij.grinberg@drexel.edu} % ENTER YOUR EMAIL HERE
\newcommand{\psetnumber}{1} % ENTER THE NUMBER OF THIS PSET HERE

%----------------------------------------------------------------------------------------
%	HEADER AND FOOTER
%----------------------------------------------------------------------------------------
\ihead{Solutions to homework set \#\psetnumber} % Page header left
\ohead{page \thepage\ of \pageref{LastPage}} % Page header right
\ifoot{\myname} % left footer
\ofoot{\mymail} % right footer

%----------------------------------------------------------------------------------------
%	TITLE SECTION
%----------------------------------------------------------------------------------------
\title{	
\normalfont \normalsize 
\textsc{Drexel University, Department of Mathematics} \\ [25pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge Math 235: Mathematical Problem Solving, \\
Fall 2021:
Homework \psetnumber\\% The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}
\author{\myname}

\begin{document}

\maketitle % This command causes the title part to be printed.

%----------------------------------------------------------------------------------------
%	EXERCISE 1
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 1}

\subsection{Problem}

Let $n$ be a positive integer.
For each $k \in \set{1, 2, \ldots, n-1}$, we let
\[
a_k := \tup{n-k} \prod_{i=0}^{k-2} \tup{n-i}
     = \tup{n-k} \tup{n-k+2} \tup{n-k+3} \cdots n .
\]
Prove that
\[
\sum_{k=1}^{n-1} a_k = n! - 1.
\]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 2
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 2}

\subsection{Problem}

Let $n \in \NN$. Prove that
\[
\sum_{r=1}^{n} \dfrac{1}{r} \dbinom{n}{r}
= \sum_{r=1}^{n} \dfrac{2^r - 1}{r}.
\]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 3
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 3}

\subsection{Problem}

Let $n$ and $p$ be positive integers such that $p \leq 2n$. Prove that
\[
\sum_{k=p}^{n} 2^k k \dbinom{2n-k-1}{n-1}
= 2^p n \dbinom{2n-p}{n}.
\]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 4
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 4}

\subsection{Problem}

A country has $n$ towns (with $n\geq 1$), arranged along a
linear road running from left to right. Each town has a \emph{left bulldozer}
(standing on the road to the left of the town and facing left) and a
\emph{right bulldozer} (standing on the road to the right of the town and
facing right). The sizes of the $2n$ bulldozers are distinct. Every time when
a left and right bulldozer confront each other, the larger bulldozer pushes
the smaller one off the road. On the other hand, bulldozers are unprotected at
their rears; so, if a bulldozer reaches the rear-end of another one, the first
one pushes the second one off the road, regardless of their sizes.

For any two towns $P$ and $Q$, we say that town $P$ \emph{dominates} town $Q$
if the bulldozer of $P$ that is facing in the direction of $Q$ can move over
to $Q$ without getting pushed off the road.

Prove that there is exactly one town that is not dominated by any other town.

[\textbf{Example:} Here is one possibility for $n=5$:
\[
\underline{\ \ \ \ \ 7A3\ \ \ \ \ 5B4\ \ \ \ \ 1C9\ \ \ \ \ 2D6\ \ \ \ \ 8E10\ \ \ \ \ }
\ ,
\]
where $A,B,C,D,E$ are the five towns and where each number stands for the size
of the corresponding bulldozer. It is easy to check that in this
configuration, town $A$ dominates no other town; town $B$ dominates towns $A$,
$C$ and $D$; town $C$ dominates $D$ and $E$; town $D$ dominates no other
towns; town $E$ dominates town $D$. Thus, the unique undominated town is $B$.]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 5
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 5}

\subsection{Problem}

Let $k$ and $n$ be two nonnegative integers.
Let $S$ be a set with size $\abs{S} \geq k \tup{n+1} - 1$.
Assume that each $n$-element subset of $S$ is colored either
red or green.
Prove that there exist $k$ pairwise disjoint $n$-element
subsets of $S$ that have the same color.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 6
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 6}

\subsection{Problem}

Let $n$ be a positive integer. You have $n$ boxes, and each
box contains a non-negative number of pebbles. In each move, you are allowed
to take two pebbles from an arbitrary box, throw away one of the pebbles and
put the other pebble in another box. (You can freely choose these two boxes.)
An initial configuration of pebbles is called \emph{solvable} if it is
possible to reach a configuration with no empty box, in a finite (possibly
zero) number of moves.

Prove that a configuration $\tup{a_1, a_2, \ldots, a_n}$ (that
is, a configuration in which box $1$ has $a_1$ pebbles, box $2$ has $a_2$
pebbles, and so on) is solvable if and only if
\[
  \left\lceil \dfrac{a_1}{2} \right\rceil
+ \left\lceil \dfrac{a_2}{2} \right\rceil
+ \cdots
+ \left\lceil \dfrac{a_n}{2} \right\rceil \geq n.
\]
Here, $\left\lceil x \right\rceil$ denotes the \emph{ceiling} of a real number
$x$ (that is, the smallest integer that is $\geq x$).

[\textbf{Example:} The configuration $\tup{3,0,6,0,0}$ is solvable.
Indeed, we can take two pebbles from the first box and move one of them to the
second, obtaining $\tup{1,1,6,0,0}$; then we can take two pebbles
from the third box and move one of them to the fourth, obtaining
$\tup{1,1,4,1,0}$; and finally take two pebbles from the third box and move
one of them to the fifth, obtaining $\tup{1,1,2,1,1}$.]

\subsection{Solution}

...



\horrule{0.3pt} \\[0.4cm]

\begin{thebibliography}{99999999}                                                                                         %

\end{thebibliography}

\end{document}

