408 lines
14 KiB
TeX
Executable File
408 lines
14 KiB
TeX
Executable File
% use [nosolutions] flag to hide solutions.
|
|
% use [solutions] flag to show solutions.
|
|
\documentclass[
|
|
solutions
|
|
]{../../resources/ormc_handout}
|
|
\usepackage{../../resources/macros}
|
|
|
|
\usepackage{amsmath}
|
|
\usepackage{amssymb}
|
|
\usepackage{tikz}
|
|
|
|
\uptitlel{Advanced 2}
|
|
\uptitler{\smallurl{}}
|
|
\title{Pidgeonhole Problems}
|
|
\subtitle{Prepared by Mark on \today}
|
|
|
|
\begin{document}
|
|
|
|
\maketitle
|
|
|
|
|
|
\problem{}
|
|
Is it possible to cover an equilateral triangle with two smaller equilateral triangles? Why or why not?
|
|
|
|
|
|
\begin{solution}
|
|
In order to completely cover an equilateral triangle, the two smaller triangles must cover all three vertices. Since the longest length of an equilateral triangle is one of its sides, a smaller triangle cannot cover more than one vertex. Therefore, we cannot completely cover the triangle with two smaller copies. \par
|
|
|
|
\textcolor{gray}{\textit{Bonus question:}} Can you cover a square with three smaller squares?
|
|
\end{solution}
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
|
|
\problem{}<divisibledifference>
|
|
You are given $n + 1$ distinct integers. \par
|
|
Prove that at least two of them have a difference divisible by $n$.
|
|
|
|
\begin{solution}
|
|
$n~|~(a-b) \iff a \equiv b \pmod{n}$ \par
|
|
|
|
Let $i_0 ... i_{n+1}$ be our set of integers. If we pick $i_0 ... i_{n+1}$ so that no two have a difference divisible by $n$, we must have $i_0 \not\equiv i_k \pmod{n}$ for all $1 \leq k \leq n+1$. There are $n$ such $i_k$, and there are $n$ equivalence classes mod $n$. \par
|
|
|
|
Therefore, either, $i_1 ... i_{n+1}$ must cover all equivalence classes mod $n$ (implying that $i_0 \equiv i_k \pmod{n}$ for some k), or there exist two elements in $i_1 ... i_{n+1}$ that are equivalent mod $n$. \par
|
|
|
|
In either case, we can find $a, b$ so that $a \equiv b \pmod{n}$, which implies that $n$ divides $a-b$.
|
|
\end{solution}
|
|
|
|
\vfill
|
|
\pagebreak
|
|
|
|
|
|
|
|
|
|
\problem{}
|
|
You have an $8 \times 8$ chess board with two opposing corner squares cut off. You also have a set of dominoes, each of which is the size of two squares. Is it possible to completely cover the board with dominos, so that none overlap nor stick out?
|
|
|
|
\begin{solution}
|
|
A domino covers two adjacent squares. Adjacent squares have different colors. \par
|
|
|
|
If you remove two opposing corners of a chessboard, you remove two squares of the same color, and you're left with $32$ of one and $30$ of the other. \par
|
|
|
|
Since each domino must cover two colors, you cannot cover the modified board.
|
|
\end{solution}
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
\problem{}
|
|
The ocean covers more than a half of the Earth's surface. Prove that the ocean has at least one pair of antipodal points.
|
|
|
|
\begin{solution}
|
|
Let $W$ be the set of wet points, and $W^c$ the set of points antipodal to those in $W$. $W$ and $W^c$ each contain more than half of the points on the earth. The set of dry points, $D$, contains less than half of the points on the earth. Therefore, $W^c \not\subseteq D$. \par
|
|
|
|
\textcolor{gray}{\textit{Note:}} This solution isn't very convincing. However, it is unlikely that the students know enough to provide a fully rigorous proof.
|
|
\end{solution}
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
\problem{}
|
|
There are $n > 1$ people at a party. Prove that among them there are at least two people who have the same number of acquaintances at the gathering. (We assume that if A knows B, then B also knows A)
|
|
|
|
\begin{solution}
|
|
Assume that every attendee knows a different number of people. There is only one way this may happen: the most popular person knows $n-1$ people (that is, everyone but himself), the second-most popular knows $n-2$, etc. The least-popular person must then know $0$ people. \par
|
|
|
|
This is impossible, since we know that someone must know $n-1$. \par
|
|
(Remember, ``knowing'' must be mutual.)
|
|
\end{solution}
|
|
|
|
\vfill
|
|
\pagebreak
|
|
|
|
|
|
|
|
\problem{}
|
|
Pick five points in $\mathbb{R}^2$ with integral coordinates. Show that two of these form a line segment that has an integral midpoint.
|
|
|
|
\begin{solution}
|
|
Let $e, o$ represent even and odd integers. \par
|
|
There are four possible classes of points: $(e, e)$, $(o, o)$, $(e, o)$, $(o, e)$. \par
|
|
|
|
$\text{midpoint}(a, b) = (\frac{a_x + b_x}{2}, \frac{a_y + b_y}{2})$. If $a_x + b_x$ and $a_y + b_y$ are both even, the midpoint of points $a$ and $b$ will have integer coordinates. \par
|
|
|
|
Since we pick five points from four classes, at least two must come from the same class. \par
|
|
$e + e = e$ and $o + o = e$, so the midpoint between two points of the same class must have integral coordinates. \par
|
|
|
|
\end{solution}
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
|
|
\problem{}<line_threecolor>
|
|
Every point on a line is painted black or white. Show that there exist three points of the same color where one is the midpoint of the line segment formed by the other two.
|
|
|
|
\begin{solution}
|
|
This is a proof by contradiction. We will try to construct a set of points where three points have such an arrangement. \par
|
|
|
|
We know that some two points on the line will have the same color:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}
|
|
% Axis
|
|
\draw[latex-latex] (-3,0) -- (4,0);
|
|
|
|
% Ticks
|
|
\foreach \x in {-2,-1,0,1,2,3}
|
|
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
|
|
|
|
% Points
|
|
\path [draw=black, fill=black] (1,0) circle (2pt);
|
|
\path [draw=black, fill=black] (0,0) circle (2pt);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
This implies that the points one unit left and right of them must also be white---if they are not, they will form a line of equidistant black points.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}
|
|
% Axis
|
|
\draw[latex-latex] (-3,0) -- (4,0);
|
|
|
|
% Ticks
|
|
\foreach \x in {-2,-1,0,1,2,3}
|
|
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
|
|
|
|
% Points
|
|
\path [draw=black, fill=black] (1,0) circle (2pt);
|
|
\path [draw=black, fill=black] (0,0) circle (2pt);
|
|
\path [draw=black, fill=white] (2,0) circle (2pt);
|
|
\path [draw=black, fill=white] (-1,0) circle (2pt);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Our original assumption also implies that the center point is white. \par
|
|
This, however, creates a line of equidistant white points:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}
|
|
% Axis
|
|
\draw[latex-latex] (-3,0) -- (4,0);
|
|
|
|
% Ticks
|
|
\foreach \x in {-2,-1,0,1,2,3}
|
|
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
|
|
|
|
% Points
|
|
\path [draw=black, fill=black] (1,0) circle (2pt);
|
|
\path [draw=black, fill=black] (0,0) circle (2pt);
|
|
\path [draw=black, fill=white] (0.5,0) circle (2pt);
|
|
\path [draw=black, fill=white] (2,0) circle (2pt);
|
|
\path [draw=black, fill=white] (-1,0) circle (2pt);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
It is thus impossible to create a set of points that does not have the property stated in the problem.
|
|
\end{solution}
|
|
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
\problem{}
|
|
Every point on a plane is painted black or white. Show that there exist two points in the plane that have the same color and are located exactly one foot away from each other.
|
|
|
|
\begin{solution}
|
|
Pick three points that form an equilateral triangle with side length 1.
|
|
\end{solution}
|
|
|
|
\vfill
|
|
\pagebreak
|
|
|
|
|
|
|
|
\problem{}<multipleofones>
|
|
Let n be an integer not divisible by $2$ and $5$. Show that n has a multiple consisting entirely of ones.
|
|
|
|
\begin{solution}
|
|
Let $a_1 = 1, a_2 = 11, a_3 = 111$, and so on.
|
|
|
|
\vspace{2mm}
|
|
|
|
Consider the sequence $a_1, ..., a_{n+1}$. \par
|
|
By \ref{divisibledifference}, there exist $a_i$ and $a_j$ in this list so that $a_i - a_j \equiv 0 \pmod{n}$. \par
|
|
|
|
\vspace{2mm}
|
|
|
|
Since $a_i$ and $a_j$ are both made of ones, $a_i - a_j = 11...111 \times 10^j$. \par
|
|
$n$ must be a factor of either $11...111$ or $10^j$. \par
|
|
Since $n$ isn't divisible by $2$ or $5$, $10^j$ cannot be divisible by $n$, so $11...111$ must be a factor of $n$.
|
|
\end{solution}
|
|
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
|
|
\problem{}
|
|
Prove that for any $n > 1$, there exists an integer made of only sevens and zeros that is divisible by $n$.
|
|
|
|
|
|
\begin{solution}
|
|
If $n$ is not divisible by $2$ or $5$, the solution to this problem is the same as \ref{multipleofones}: \par
|
|
just multiply the number of all ones by $7$.
|
|
|
|
\vspace{2mm}
|
|
|
|
If $n$ is divisible by $2$ or $5$, set $p$ to the largest power of $2$ or $5$ in $n$. \par
|
|
Multiply the above number by $10^p$ to get a number that satisfies the conditions above.
|
|
\end{solution}
|
|
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
\problem{}
|
|
Choose $n + 1$ integers between $1$ and $2n$. Show that at least two of these are co-prime.
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
\problem{}
|
|
Choose $n + 1$ integers between $1$ and $2n$. Show that you must select two numbers $a$ and $b$ such that $a$ divides $b$.
|
|
|
|
\begin{solution}
|
|
Split the set $\{1, ..., 2n\}$ into classes defined by each integer's greatest odd divisor. There will be $n$ classes since there are $\frac{k}{2}$ odd numbers between $1$ and $n$. Because we pick $n + 1$ numbers, at least two will come from the same class---they will be divisible. \par
|
|
|
|
For example, if $n = 5$, our classes are
|
|
\begin{itemize}
|
|
\item[1:] $\{1, 2, 4, 8\}$
|
|
\item[3:] $\{3, 6\}$
|
|
\item[5:] $\{5, 10\}$
|
|
\item[7:] $\{7\}$
|
|
\item[9:] $\{9\}$
|
|
\end{itemize}
|
|
\end{solution}
|
|
|
|
\vfill
|
|
\pagebreak
|
|
|
|
|
|
|
|
\problem{}
|
|
Show that it is always possible to choose a subset of the set of integers $\{a_1, a_2, ... , a_n\}$ so that the sum of the numbers in the subset is divisible by $n$.
|
|
|
|
\begin{solution}
|
|
Let $\{a_1^\prime, a_2^\prime, ..., a_n^\prime\}$ be this set mod $n$. \par
|
|
If any $a_i^\prime$ is zero, we're done: $\{a_i^\prime\}$ satisfies the problem.
|
|
|
|
\vspace{2mm}
|
|
|
|
If none are zero, consider the set $\{a_1^\prime,~ a_1^\prime + a_2^\prime,~ ...,~ a_1^\prime + a_2^\prime + ... + a_n^\prime\}$. \par
|
|
If any element of this set is zero, we're done.
|
|
|
|
\vspace{2mm}
|
|
|
|
If zero is not in this set, we have $n$ numbers with $n-1$ possible remainders. Therefore, at least two elements in this set must be equivalent mod $n$. If we subtract these two elements, we get a sum divisible by $n$.
|
|
\end{solution}
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
|
|
\problem{}
|
|
Show that there exists a positive integer divisible by $2013$ that has $2014$ as its last four digits.
|
|
|
|
\begin{solution}
|
|
Let $n$ be this number. \par
|
|
First, note that $n - 2013$ has $0001$ as its last four digits. \par
|
|
|
|
\vspace{2mm}
|
|
|
|
So, we see that $n - 2013 = 2013k \equiv 1 \pmod{1000}$. \par
|
|
Of course, $k$ = $2013^{-1} \pmod{1000}$, which exists because 2013 and 1000 are coprime. \par
|
|
And finally, we see that $n = 2013 \times (k + 1)$.
|
|
\end{solution}
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
|
|
\problem{}
|
|
Let $n$ be an odd number. Let $a_1, a_2, ... , a_n$ be a permutation of the numbers $1, 2, ... , n$. \par
|
|
Show that $(a_1 - 1) \times (a_2 - 2) \times ... \times (a_n - n)$ is even.
|
|
|
|
\begin{solution}
|
|
If $n$ is odd, there will be $m$ even and $m + 1$ odd numbers between $1$ and $n$. \par
|
|
Therefore, if we match each $a_n$ with an integer in $[1, ..., n]$, we will have to match at least one odd number with an odd number. \par
|
|
|
|
The difference of two odd numbers is even, so the product above will have at least one factor of two.
|
|
\end{solution}
|
|
|
|
\vfill
|
|
\pagebreak
|
|
|
|
|
|
|
|
|
|
\problem{}
|
|
A stressed-out student consumes at least one espresso every day of a particular year, drinking $500$ overall. Show the student drinks exactly $100$ espressos on some consecutive sequence of days.
|
|
|
|
\begin{solution}
|
|
Rearrange the problem. Don't think about days, think about espressos. Consider the following picture:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}
|
|
% Axis
|
|
\draw[-] (1,0) -- (12,0);
|
|
|
|
% Ticks
|
|
\foreach \x in {1, 2, 3, 4}
|
|
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
|
|
|
|
% Legend
|
|
\node[above] at (-0.5, 3pt) {Day consumed};
|
|
\node[below] at (-0.5,-3pt) {Espresso \#};
|
|
|
|
% Bottom numbers
|
|
\foreach \x in {1, 2, ..., 7}
|
|
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt, -3pt) node[below] {$\x$};
|
|
\draw[shift={(9,0)},color=black] (0pt,0pt) -- (0pt, -3pt) node[below] {$...$};
|
|
\draw[shift={(11,0)},color=black] (0pt,0pt) -- (0pt, -3pt) node[below] {$499$};
|
|
\draw[shift={(12,0)},color=black] (0pt,0pt) -- (0pt, -3pt) node[below] {$500$};
|
|
|
|
% Top numbers
|
|
\draw[shift={(1,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$1$};
|
|
\draw[shift={(2,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$1$};
|
|
\draw[shift={(3,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$2$};
|
|
\draw[shift={(4,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$2$};
|
|
\draw[shift={(5,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$3$};
|
|
\draw[shift={(6,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$3$};
|
|
\draw[shift={(7,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$3$};
|
|
\draw[shift={(9,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$...$};
|
|
\draw[shift={(11,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$365$};
|
|
\draw[shift={(12,0)},color=black] (0pt,0pt) -- (0pt,3pt) node[above] {$365$};
|
|
|
|
\draw[shift={(3, -1)}, color=orange, thick] (0pt,0pt) -- (0pt,3pt)
|
|
node[below, color=gray, shift={(0, -3pt)}] {$3$};
|
|
\draw[color=orange, thick] (3, -1) -- (8.5, -1) node[below, midway] {100 espressos};
|
|
\draw[shift={(8.5, -1)}, color=orange, thick] (0pt,0pt) -- (0pt,3pt)
|
|
node[below, color=gray, shift={(0, -3pt)}] {$102$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
If there exists a sequence of days where the student drinks exactly 100 espressos, we must have at least one ``block'' (in orange, above) of 100 espressos that both begins and ends on a ``clean break'' between days. \par
|
|
|
|
There are $499$ ``breaks'' between $500$ espressos. \par
|
|
In a year, there are $364$ clean breaks. This leaves $499 - 364 = 135$ ``dirty'' breaks. \par
|
|
We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \par
|
|
|
|
% spell:off
|
|
However, there are $401$ possible blocks,
|
|
since we can start one at the $1^{\text{st}},2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
|
|
% spell:on
|
|
|
|
Out of $401$ blocks, a maximum of $270$ can be dirty. We are therefore guaranteed at least $131$ clean blocks. This completes the problem---each clean block represents a set of consecutive, whole days during which exactly 100 espressos were consumed.
|
|
|
|
\end{solution}
|
|
|
|
|
|
\vfill
|
|
|
|
|
|
\problem{}
|
|
Show that there are either three mutual acquaintances or four mutual strangers at a party with ten or more people.
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
\problem{}
|
|
Given a table with a marked point, $O$, and with $2013$ properly working watches put down on the table, prove that there exists a moment in time when the sum of the distances from $O$ to the watches' centers is less than the sum of the distances from $O$ to the tips of the watches' minute hands.
|
|
|
|
\vfill
|
|
\end{document}
|