From 8ddfc5b3462eea32fe973d10053191425f61b52a Mon Sep 17 00:00:00 2001 From: Mark Date: Mon, 29 May 2023 21:49:04 -0700 Subject: [PATCH] Removed files that don't belong --- Problems/all.tex | 112 -------- Problems/problemlib.sty | 85 ------ Problems/problems/algebra.tex | 77 ------ Problems/problems/combinatorics.tex | 42 --- Problems/problems/geometry.tex | 39 --- Problems/problems/misc.tex | 245 ------------------ Problems/problems/numbertheory.tex | 172 ------------- Problems/problems/pidgeonhole.tex | 385 ---------------------------- 8 files changed, 1157 deletions(-) delete mode 100755 Problems/all.tex delete mode 100755 Problems/problemlib.sty delete mode 100755 Problems/problems/algebra.tex delete mode 100755 Problems/problems/combinatorics.tex delete mode 100755 Problems/problems/geometry.tex delete mode 100755 Problems/problems/misc.tex delete mode 100755 Problems/problems/numbertheory.tex delete mode 100755 Problems/problems/pidgeonhole.tex diff --git a/Problems/all.tex b/Problems/all.tex deleted file mode 100755 index 2e8546d..0000000 --- a/Problems/all.tex +++ /dev/null @@ -1,112 +0,0 @@ -\documentclass[ - solutions, - hidewarning, -]{../resources/ormc_handout} - - -\def\stars#1{ - \count255=1 - \ifnum #1 > 0 - \loop - {\color{cyan!50!gray} \bigstar}% - \ifnum\count255 < #1 - \advance\count255 by 1 - \repeat - \else% - {\color{cyan!50!gray} \bigstar}% - \fi% - % - \ifnum \count255 < 5 - \loop - {\color{gray} \bigstar}% - \ifnum\count255 < 4 - \advance\count255 by 1 - \repeat - \fi -} - - -\NewDocumentCommand{\p}{ m m }{ - - \begin{minipage}{\textwidth} - \textbf{#1 #2:} \par - - \ifdifficulty{#1}{#2}{{\color{gray}Difficulty:} \stars{\getdifficulty{#1}{#2}}} - - \vspace{2mm} - - \getproblem{#1}{#2} - - \ifsolution{#1}{#2}{}[ - \ifanswer{#1}{#2}{% - \vspace{2mm}% - {\color{red}\textbf{Solution:}} \getanswer{#1}{#2} - } - ] - \end{minipage} - - \ifsolution{#1}{#2}{ - \begin{solution} - \getsolution{#1}{#2} - \end{solution} - } - - \penalty-500 - - \vspace{15mm} - \vfill -} - -\newcommand{\postproblem}[2]{ - \ifcsname#1:all\endcsname - \expandafter\edef\csname#1:all\endcsname{\csname#1:all\endcsname \p{#1}{#2}} - \else - \expandafter\gdef\csname#1:all\endcsname{\p{#1}{#2}} - \fi -} - -\usepackage{problemlib} - -\input{problems/numbertheory} -\input{problems/algebra} -\input{problems/combinatorics} -\input{problems/pidgeonhole} -\input{problems/misc} -\input{problems/geometry} - -%\makeatletter -%\newcommand{\shoppinglist}[1]{Shopping list: #1\checknextarg} -%\newcommand{\checknextarg}{\@ifnextchar\bgroup{\gobblenextarg}{ and that's all!}} -%\newcommand{\gobblenextarg}[1]{ and also #1\@ifnextchar\bgroup%{\gobblenextarg}{ and that's all!}} -%\makeatother - -\newcommand{\q}[1]{ - \section{#1} - \csname#1:all\endcsname - - \pagebreak -} - - -\title{Mark's Problem Library} -\subtitle{ - This document lists all problems in this library. \\ - Use it to find problems or debug the source. -} - -\begin{document} - \maketitle - {Mark's Problem Library} - { - This document lists all problems in this library. \\ - Use it to find problems or debug the source. - } - - - \q{Algebra} - \q{Combinatorics} - \q{NumberTheory} - \q{Pidgeonhole} - \q{Geometry} - \q{Misc} -\end{document} \ No newline at end of file diff --git a/Problems/problemlib.sty b/Problems/problemlib.sty deleted file mode 100755 index 42c4fda..0000000 --- a/Problems/problemlib.sty +++ /dev/null @@ -1,85 +0,0 @@ -\NeedsTeXFormat{LaTeX2e} -\ProvidesPackage{problemlib}[2023/05/17 ORMC Problem Library] - - -% Defines a new problem. -% See problem definition files for usage examples. -\long\def\problemdef#1#2#3{ - \ifcsname#1:problem:#2\endcsname - \PackageError{problemlib}{Problem #1:#2 is already defined}{} - \fi - - % Use a group to limit scope of special commands. - \begingroup - - % \gdef = \global\def - % This makes problem definitions global, and not restricted to this group. - % - % ##1: used to define parametrized macros in a parametrized macro. - % When \problemdef is expanded, ##1 becomes #1. - % - % \long\def: like \def, but allows multi-paragraph arguments - % \def\name#1{def: #1}: define a macro \name with one arg. TeX primitive. - % - % We don't use \newcommand or \renewcommand here because they will fail if - % a macro is defined/not defined. \def will always work, defining a new name - % in this group, temporarily redefining existing commands. - % - % \expandafter - % \expandafter command delays expanding a macro until its arguments have been expanded. - \long\def\difficulty##1{\expandafter\gdef\csname#1:difficulty:#2\endcsname{##1}} - \long\def\statement##1{\expandafter\gdef\csname#1:problem:#2\endcsname{##1}} - \long\def\solution##1{\expandafter\gdef\csname#1:solution:#2\endcsname{##1}} - \long\def\answer##1{\expandafter\gdef\csname#1:answer:#2\endcsname{##1}} - - #3 - - \endgroup - - % Optional hook to run after each problem definition. - % Must be defined BEFORE problems are included. - \ifdefined\postproblem - \postproblem{#1}{#2} - \fi -} - -% Gets problem data. -% Arguments: -% #1: Category -% #2: Problem id -% #3: Field to get -\NewDocumentCommand{\get}{ m m m }{% - \ifcsname#1:#3:#2\endcsname% - \csname#1:#3:#2\endcsname% - \else% - \PackageError{problemlib}{Problem #1:#2 is not defined}{} - \fi% -} -\NewDocumentCommand{\getdifficulty}{ m m }{\get{#1}{#2}{difficulty}} -\NewDocumentCommand{\getproblem}{ m m }{\get{#1}{#2}{problem}} - - -% #1: Category -% #2: Problem id -% #3: attribute to check -% #4: Show if true -% #5: Show if false -\NewDocumentCommand{\ifattribute}{ m m m m d[] }{% - \ifcsname#1:#3:#2\endcsname#4% - \else% - \IfNoValueF{#5}{#5}% - \fi% -} -\NewDocumentCommand{\ifanswer}{ m m m d[] }{\ifattribute{#1}{#2}{answer}{#3}[#4]} -\NewDocumentCommand{\ifsolution}{ m m m d[] }{\ifattribute{#1}{#2}{solution}{#3}[#4]} -\NewDocumentCommand{\ifdifficulty}{ m m m d[] }{\ifattribute{#1}{#2}{difficulty}{#3}[#4]} - -\NewDocumentCommand{\getanswer}{ m m }{% - \ifanswer{#1}{#2}{\get{#1}{#2}{answer}}% - [\PackageError{problemlib}{Problem #1:#2 has no answer}{}]% -} - -\NewDocumentCommand{\getsolution}{ m m }{ - \ifsolution{#1}{#2}{\get{#1}{#2}{solution}}% - [\PackageError{problemlib}{Problem #1:#2 has no solution}{}]% -} \ No newline at end of file diff --git a/Problems/problems/algebra.tex b/Problems/problems/algebra.tex deleted file mode 100755 index f9d26ad..0000000 --- a/Problems/problems/algebra.tex +++ /dev/null @@ -1,77 +0,0 @@ -\problemdef{Algebra}{1}{ - \statement{ - Evaluate - $ - \bigl( 1 - \frac{1}{4} \bigr) - \bigl( 1 - \frac{1}{9} \bigr) - \bigl( 1 - \frac{1}{16} \bigr) - ~ ... ~ - \bigl( 1 - \frac{1}{255} \bigr) - $ - } - \answer{$\frac{8}{15}$} -} - - - -\problemdef{Algebra}{2}{ - \statement{ - $(a + b)(a + b - 1) = ab$ and $a^2 + b^2 = 3$. \par - Find $a^3 + b^3$. - } - \answer{3} -} - - - -\problemdef{Algebra}{3}{ - \statement{ - Simplify $(2^{62} + 1)/(2^{31} + 2^{16} + 1)$ - } - \answer{$2^{31} - 2^{16} + 1$} -} - - - -\problemdef{Algebra}{4}{ - \statement{ - $x, y, z > 0$ and $xyz = 1$. \par - Also, $x + 1/z = 5$ and $y + 1/x = 29$. - - Find $z + 1/y$. - } - \answer{$z + 1/y = 1/4$} -} - - - -\problemdef{Algebra}{5}{ - \statement{ - Factor $x^8 + x^4 + 1$ into four quadratics. - } - \answer{$(x^2 - \sqrt{3} x + 1)~(x^2 + \sqrt{3} x + 1)~(x^2 - x + 1)~(x^2 + x + 1)$} -} - -\problemdef{Algebra}{6}{ - \statement{ - Sophia bought a Greyhound ticket, but then her plans changed and she sold it back for $\$24$. The percent of the ticket's cost that she lost in the sale is equal to the dollar value of her initial ticket. How much did she buy it for? List all options. - } - - \answer{$\$40$ or $\$60$} -} - -\problemdef{Algebra}{7}{ - \statement{ - How do you cut a cake into 6 pieces so that it can be distributed equally to both three guests and four guests? - } - - \answer{$\frac{3/12} + \frac{3}{4}$ or $\frac{2}{12} + \frac{2}{6} + \frac{2}{4}$} -} - -\problemdef{Algebra}{8}{ - \statement{ - On the first day the grocery store sold $\frac{1}{2}$ of all the geese and half a goose, on the second --- $\frac{1}{3}$ of the remainder and another $\frac{1}{3}$ of the goose, on the third --- $\frac{1}{4}$ of the new remnant and another 3/4 of the goose, on the fourth --- $\frac{1}{5}$ of the remainder and another $\frac{1}{5}$ of the goose. On the fifth day, the store sold the remaining 19 geese. How many geese were there in the store? - } - - \answer{101} -} \ No newline at end of file diff --git a/Problems/problems/combinatorics.tex b/Problems/problems/combinatorics.tex deleted file mode 100755 index 8170d63..0000000 --- a/Problems/problems/combinatorics.tex +++ /dev/null @@ -1,42 +0,0 @@ -\problemdef{Combinatorics}{1}{ - \statement{ - A real estate investor asked a developer to paint 2017 houses so that at least 1000 are green and 1000 are red. What is the maximum number of colors the developer can use? - } - - \answer{19} -} - -\problemdef{Combinatorics}{2}{ - \statement{ - How many ways are there to cut one $3 \times 5$ rectangle into five $1 \times 3$ rectangles? - } - - \answer{4} -} - -\problemdef{Combinatorics}{3}{ - \statement{ - How many different integral solutions $(x, y, z)$ are there to $x+y+z = 20$? - } - - \answer{$C_{19}^2$} -} - - -\problemdef{Combinatorics}{4}{ - \statement{ - Consider an uncolored $8 \times 8$ board. How many ways are there to paint the squares black or white so that we end up with exactly 31 black squares, none of which share an edge? - } - - \answer{68} -} - -\problemdef{Combinatorics}{5}{ - \statement{ - Let $A$ be the set of four-digit integers in which the first digit is equal to the sum of the other three. \par - Let $B$ be the set of four-digit integers in which the last digit is equal to the sum of the other three. \par - Which set is larger, and by how many elements? - } - - \answer{$A$, by 54.} -} \ No newline at end of file diff --git a/Problems/problems/geometry.tex b/Problems/problems/geometry.tex deleted file mode 100755 index 6fa6e3b..0000000 --- a/Problems/problems/geometry.tex +++ /dev/null @@ -1,39 +0,0 @@ -\problemdef{Geometry}{1}{ - \statement{ - What is the largest number of obtuse angles that nine rays emanating from one point can form on a plane? - } - - \answer{27} -} - -\problemdef{Geometry}{2}{ - \statement{ - The three sides of the trapezoid are equal, and one of the angles is $40^\circ$ larger than the other. Find an acute angle between the diagonals of the trapezoid. - } - - \answer{$70^\circ$} -} - -\problemdef{Geometry}{3}{ - \statement{ - In a cyclic quadrilateral $ABCD$ angle $ABD$ is $30^\circ$ and angle $BCA$ is $42^\circ$. Find angle $DAB$. - } - - \answer{$108^\circ$} -} - -\problemdef{Geometry}{4}{ - \statement{ - In a triangle $ABC$, angle $B$ is $20^\circ$, and angle $C$ is $40^\circ$. The bisector $AD$ of angle $A$ is 12. Find the difference between the sides $BC$ and $AB$. - } - - \answer{12} -} - -\problemdef{Geometry}{5}{ - \statement{ - What is the largest number of sides a polygon that is the intersection of a quadrilateral and a triangle can have? - } - - \answer{8} -} \ No newline at end of file diff --git a/Problems/problems/misc.tex b/Problems/problems/misc.tex deleted file mode 100755 index cf81af9..0000000 --- a/Problems/problems/misc.tex +++ /dev/null @@ -1,245 +0,0 @@ -\problemdef{Misc}{1}{ - \difficulty{1} - - \statement{ - A carpenter cut a chessboard into $1 \times 1$ squares in 70 minutes. \par - How long will it take him to cut it into $2 \times 2$ squares? - } - - \answer{30} -} - - -\problemdef{Misc}{2}{ - \difficulty{2} - - \statement{ - There are two kinds of books on a shelf: those on permissible magic and those on black magic. Two books on permissible magic cannot be set between exactly three other books, and two books on black magic may not stand next to each other. \par - What is the maximal amount of books that may be placed on the shelf? - } - - \answer{8} -} - - -\problemdef{Misc}{3}{ - \difficulty{3} - - \statement{ - The numbers $1 ... 9$ are arranged in a $3 \times 3$ grid. \par - The sum of each row and column is then computed. \par - What is the maximum number of consecutive integers one may find in the set of these sums? - } - - \answer{5} -} - - -\problemdef{Misc}{4}{ - \difficulty{4} - - \statement{ - 16 rugby teams participate in a regional championship. Each pair of teams plays against each other twice. The 8 teams with the most wins will proceed to the national championship. If there is a tie in this ranking, the tied teams will draw lots. \par - Assume a rugby game can never tie. What is the minimum number of wins a team needs to guarantee a spot in the nationals? - } - - \answer{23} -} - - - - -\problemdef{Misc}{5}{ - \difficulty{5} - - \statement{ - Five boxes are filled with pastries. We know that box C contains a third of the pastries in E, and that B contains two times more than C and E combined. A contains half the number of pastries in E, and a tenth of those in D. Box B contains four times more pastries than D. \par - What is the minimal possible positive number of pastries in all the boxes put together? - } - - \answer{310} -} - - -\problemdef{Misc}{6}{ - \statement{ - All faces of the cube are painted in different colors (each face is painted with the same color). If you look at this cube from one side, then you can see the blue, white and yellow faces, on the other side you can see the black, blue and red faces, and on the third side you can see the green, black and white faces. Which face is opposite to the white one? - } - - \answer{Red} -} - -\problemdef{Misc}{7}{ - \statement{ - A military band was performing in a field. First, the musicians lined up in a square. Then they rearranged themselves into a rectangle, and the number of ranks increased by 5. How many musicians were in the orchestra? - } - - \answer{400} -} - - -\problemdef{Misc}{8}{ - \statement{ - Out of five coins, three coins are real and two are fake. Both fake coins weigh the same, but each weighs less than a real coin. What is the least number of times you have to use the balance scale to guarantee that you can find at least one real coin? - } - - \answer{1} -} - - -\problemdef{Misc}{9}{ - \statement{ - Out of 81 coins one is fake and 80 are genuine. A fake coin weighs less than a genuine coin. What is the least number of times you can use a balance scale to find the fake coin? - } - - \answer{4} -} - - -\problemdef{Misc}{10}{ - \statement{ - The numbers $1, 2, 3, ..., 99, 100$ are written on the board. It is allowed to erase any two numbers a and b and write the number $a + b - 1$ instead. What number can remain on the board after 99 such operations? - } - - \answer{4951} -} - - -\problemdef{Misc}{11}{ - \statement{ - Numbers from 1 to 2021 are written on the board. With one operation it is allowed to erase two numbers and write their positive difference instead. At some point there will be only one number left on the board. What is the minimal possible value of this number? - } - - \answer{1} -} - - -\problemdef{Misc}{12}{ - \statement{ - On the island of knights, knaves and spies, you come across three people. One wears blue, one wears red, and one wears green. You know that one is a knight, one is a knave, and one is a spy. ``Who is the spy?'' you ask. \par - - The man wearing blue says, ``That man in red is the spy.'' \par - The man wearing red says, ``No, the man in green is the spy.'' \par - The man wearing green says, ``No, the man in red is in fact the spy. \par - - Knights always tell the truth, knaves always lie, and spies do whatever they want. Who is the spy? Who is the knight and who is the knave? - } - - \answer{Blue is knave, Red is knight, Green is spy.} -} - -\problemdef{Misc}{13}{ - % Induction - \statement{ - Find a formula for $1 + 2 + ... + n$. - } - - \answer{$\frac{n(n+1)}{2}$} -} - -\problemdef{Misc}{14}{ - % Induction - \statement{ - Find a formula for $(1 \times 2) + (2 \times 3) +... + (n \times (n + 1))$. - } - - \answer{$\frac{n(n+1)(n+2)}{3}$} -} - -\problemdef{Misc}{15}{ - % Induction - \statement{ - $x + \frac{1}{x} = 10$. Find $x^2 + \frac{1}{x^2}$, $x^3 + \frac{1}{x^3}$, and $x^4 + \frac{1}{x^4}$ - } - - \answer{98, 970, and 9602} -} - - -\problemdef{Misc}{16}{ - % Induction - \statement{ - Let $F_n$ be the $n$th Fibonacci number. $F_0 = 0, F_1 = 1$. \par - How many of $F_0, F_1, ..., F_{2021}$ are even? - } - - \answer{674} -} - -\problemdef{Misc}{17}{ - % Induction - \statement{ - How many ways are there to split a $2 \times 10$ into $1 \times 2$ or $2 \times 1$ dominoes? - } - - \answer{89} -} - -\problemdef{Misc}{18}{ - \statement{ - Split the shape below into six equal parts. \par - Each part must have the same shape, but they may be rotated. - - \begin{center} - \begin{tikzpicture}[scale = 0.4] - \foreach \x in {0, ..., 7} { - \foreach \y in {0, ..., 7} { - \draw[draw=black] (\x,\y) rectangle +(1,1); - }} - - \draw[draw=black] (0,8) rectangle +(1,1); - \draw[draw=black] (1,8) rectangle +(1,1); - \draw[draw=black] (-1,1) rectangle +(1,1); - \draw[draw=black] (-1,0) rectangle +(1,1); - \draw[draw=black] (6,-1) rectangle +(1,1); - \draw[draw=black] (7,-1) rectangle +(1,1); - \draw[draw=black] (8,6) rectangle +(1,1); - \draw[draw=black] (8,7) rectangle +(1,1); - \end{tikzpicture} - \end{center} - } - - \solution{ - \begin{center} - \begin{tikzpicture}[scale = 0.4] - \fill[yellow] (-1,0) rectangle +(4,2); - \fill[yellow] (0,1) rectangle +(1,4); - \fill[yellow] (0,4) rectangle +(2,1); - - \fill[red] (6,-1) rectangle +(2,4); - \fill[red] (3,0) rectangle +(4,1); - \fill[red] (3,0) rectangle +(1,2); - - \fill[blue] (4,1) rectangle +(2,4); - \fill[blue] (1,2) rectangle +(4,1); - \fill[blue] (1,2) rectangle +(1,2); - - \fill[cyan] (2,3) rectangle +(2,4); - \fill[cyan] (3,5) rectangle +(4,1); - \fill[cyan] (6,4) rectangle +(1,2); - - \fill[green] (0,5) rectangle +(2,4); - \fill[green] (1,7) rectangle +(4,1); - \fill[green] (4,6) rectangle +(1,2); - - \fill[gray] (5,6) rectangle +(4,2); - \fill[gray] (7,3) rectangle +(1,4); - \fill[gray] (6,3) rectangle +(2,1); - - \foreach \x in {0, ..., 7} { - \foreach \y in {0, ..., 7} { - \draw[draw=black] (\x,\y) rectangle +(1,1); - }} - - \draw[draw=black] (0,8) rectangle +(1,1); - \draw[draw=black] (1,8) rectangle +(1,1); - \draw[draw=black] (-1,1) rectangle +(1,1); - \draw[draw=black] (-1,0) rectangle +(1,1); - \draw[draw=black] (6,-1) rectangle +(1,1); - \draw[draw=black] (7,-1) rectangle +(1,1); - \draw[draw=black] (8,6) rectangle +(1,1); - \draw[draw=black] (8,7) rectangle +(1,1); - \end{tikzpicture} - \end{center} - } -} \ No newline at end of file diff --git a/Problems/problems/numbertheory.tex b/Problems/problems/numbertheory.tex deleted file mode 100755 index 9b3a868..0000000 --- a/Problems/problems/numbertheory.tex +++ /dev/null @@ -1,172 +0,0 @@ -\problemdef{NumberTheory}{1}{ - - \statement{ - Starting September 1, four mathcirclers began to visit the cinema. The first visited it every fourth day, the second --- every fifth, the third --- every sixth and the fourth --- every ninth. \par - - When will all the circlers meet at the cinema for the second time? - } - - \answer{34} -} - - - -\problemdef{NumberTheory}{2}{ - % МАТЕМ + АТИКА = 187407 - \statement{ - Each letter in $MATHM + AJORS$ represents a single-digit number. Maximize this quantity. - } - \answer{UNKNOWN} -} - - - -\problemdef{NumberTheory}{3}{ - \statement{ - $Q$ is a three digit number. \par - $Q - 7$ is divisible by 7. $Q - 8$ is divisible by 8. $Q - 9$ is divisible by 9. What is $Q$? - } - - \answer{504} -} - - - -\problemdef{NumberTheory}{4}{ - - \statement{ - Alex and Anna share a tub of popcorn. Alex eats one kernel, Anna eats two. Alex then eats three, and the pattern continues. The person that takes the final turn consumes all the remaining popcorn, even if there aren't enough kernels for a complete turn. - - Alex ate 2017 kernels. How many were left for Anna? - } - - \answer{1980} -} - - - -\problemdef{NumberTheory}{5}{ - - \statement{ - Several positive integers were multiplied to get $224$. \par - The smallest of these was exactly equal to half the largest. \par - How many numbers were multiplied? - } - - \answer{3} -} - - - -\problemdef{NumberTheory}{6}{ - \statement{ - How many natural numbers $n$ less than 10,000 satisfy $2^n - n^2 \equiv 0~~\text{(mod 7)}$? - } - - \answer{2858} -} - - - -\problemdef{NumberTheory}{7}{ - \statement{ - Kolya was supposed to multiply a single-digit number and a two-digit one, but instead, he wrote them down in a row and got a three-digit number, which turned out to be three times more than the product that he was supposed to compute. \par - - What numbers could Kolya have? List all the possibilities. - } - - \answer{$7 \times 35$ or $1 \times 50$ or $2 \times 40$} -} - -\problemdef{NumberTheory}{8}{ - \statement{ - Represent the number 2021 as a sum of four positive integers so that all the digits in these numbers are different. - } - - \answer{$2021 = 1987 + 23 + 6 + 5$ Others are possible.} -} - -\problemdef{NumberTheory}{9}{ - \statement{ - Find the largest positive integer in which each internal digit is greater than half the sum of the two adjacent digits - } - - \answer{36899863} -} - -\problemdef{NumberTheory}{10}{ - \statement{ - In the cells of a $5 \times 5$ square, the numbers are arranged so that the sums of the numbers in all rows and in all columns are the same. The sum of all the numbers in the upper left $2 \times 2$ square is $10$, and in the lower right $3 \times 3$ square is $15$. Find the sum of all the numbers in the table. - } - - \answer{25} -} - - - -\problemdef{NumberTheory}{11}{ - \statement{ - Several numbers consisting only of ones (like 1 or 1111), were added together, and the result was the number 2021. Find the smallest possible number of terms. - } - - \answer{11} -} - - - -\problemdef{NumberTheory}{12}{ - \statement{ - Find the largest ten-digit number whose first digit is divisible by 1, the second by 2, $...$ , and the tenth by 10. - } - - \answer{9898567890} -} - - - -\problemdef{NumberTheory}{13}{ - \statement{ - Find any positive integers $a$ and $b$ so that the fractions $\frac{a}{b}$, $\frac{a + 1}{b}$, $\frac{a+1}{b+1}$ are irreducible. - } - - \answer{Most numbers work. Checking is easy.} -} - -\problemdef{NumberTheory}{14}{ - \statement{ - The numbers $1, 2, ..., 25$ are written out in a $5 \times 5$ table so that in each line the numbers are arranged in ascending order. What is the largest value that the sum of the numbers in the third column can take? - } - - \answer{85} -} - - -\problemdef{NumberTheory}{15}{ - \statement{ - Find the smallest natural number whose sum and product of digits are equal to 80. - } - - \answer{$11...1258$, which contains 65 ones.} -} - - -\problemdef{NumberTheory}{16}{ - \statement{ - The math teacher uses a problem book, which contains one hundred problems with numbers from 1 to 100. At the beginning of each lesson, the teacher attaches the numbers of three problems to the magnetic board. To do this, he uses magnets in form of digits. What is the smallest number of magnets a teacher needs to buy so that he can compose the numbers of any three problems from this book? (digit 6 can be flipped) - } - - \answer{38} -} - -\problemdef{NumberTheory}{17}{ - \statement{ - Arrange the following numbers in ascending order: \par - - $(-\frac{2}{3})^1$, $(-\frac{2}{3})^2$, $(-\frac{2}{3})^3$, $(-\frac{2}{3})^4$ - } - - \answer{$(-\frac{2}{3})^1 < (-\frac{2}{3})^3 < (-\frac{2}{3})^4 < (-\frac{2}{3})^2$} -} - - - diff --git a/Problems/problems/pidgeonhole.tex b/Problems/problems/pidgeonhole.tex deleted file mode 100755 index 028b068..0000000 --- a/Problems/problems/pidgeonhole.tex +++ /dev/null @@ -1,385 +0,0 @@ -\problemdef{Pidgeonhole}{1}{ - \difficulty{1} - - \statement{ - Is it possible to cover an equilateral triangle with two smaller equilateral triangles? Why or why not? - } - - \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? - } -} - - - - -\problemdef{Pidgeonhole}{2}{ - \difficulty{3} - - \statement{ - You are given $n + 1$ integers. \par - Prove that at least two of them have a difference divisible by $n$. - } - - \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 $a-b$. - } -} - - - - -\problemdef{Pidgeonhole}{3}{ - \difficulty{1} - - \statement{ - 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 the board with dominos, so that none overlap nor stick out? - } - - \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. - } -} - -\problemdef{Pidgeonhole}{4}{ - \difficulty{2} - - \statement{ - The ocean covers more than a half of the Earth's surface. Prove that the ocean has at least one pair of antipodal points. - } - - \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. - } -} - - -\problemdef{Pidgeonhole}{5}{ - \difficulty{2} - - \statement{ - 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) - } - - \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.) - } -} - - -\problemdef{Pidgeonhole}{6}{ - \difficulty{3} - - \statement{ - Pick five points in $\mathbb{R}^2$ with integral coordinates. Show that two of these form a line segment that has an integral midpoint. - } - - \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 - } -} - - -\problemdef{Pidgeonhole}{7}{ - \difficulty{2} - - \statement{ - 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. - } - - \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. - } -} - -\problemdef{Pidgeonhole}{8}{ - \difficulty{1} - - \statement{ - 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. - } - - \solution{Pick three points that form an equilateral triangle with side length 1.} -} - -\problemdef{Pidgeonhole}{9}{ - \difficulty{2} - - \statement{ - Each point on a circle is colored either black or white. Prove that there exist three equally spaced points of the same color. - } - - \solution{This problem is exactly the same as Pidgeonhole 7} -} - - -\problemdef{Pidgeonhole}{10}{ - \difficulty{4} - - \statement{ - Let n be an integer not divisible by $2$ and $5$. Show that n has a multiple consisting entirely of ones. - } -} - - - - - - -\problemdef{Pidgeonhole}{11}{ - \difficulty{5} - - \statement{ - Prove that for any $n > 1$, there exists an integer made of only sevens and zeros that is divisible by $n$. - } -} - - - - - -\problemdef{Pidgeonhole}{12}{ - \difficulty{4} - - \statement{ - Choose $n + 1$ integers between $1$ and $2n$. Show that at least two of these are co-prime. - } -} - - - - - -\problemdef{Pidgeonhole}{13}{ - \difficulty{3} - - \statement{ - Choose $n + 1$ integers between $1$ and $2n$. Show that you must select two numbers $a$ and $b$ such that $a$ divides $b$. - } - - \solution{ - Split the 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} - } -} - - -\problemdef{Pidgeonhole}{14}{ - \difficulty{4} - - \statement{ - 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 $n$. - } -} - - - - - - -\problemdef{Pidgeonhole}{15}{ - \difficulty{4} - - \statement{ - Show that there exists a positive integer divisible by $2013$ that has $2014$ as its last four digits. - } -} - - - - -\problemdef{Pidgeonhole}{16}{ - \difficulty{3} - - \statement{ - 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. - } - - \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. - } -} - - - -\problemdef{Pidgeonhole}{17}{ - \difficulty{4} - - \statement{ - 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. - } - - \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 - - However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par - - 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. - } -} - - - - - - - -\problemdef{Pidgeonhole}{18}{ - \difficulty{1} - - \statement{ - Show that there are either three mutual acquaintances or four mutual strangers at a party with ten or more people. - } -} - - - - - -\problemdef{Pidgeonhole}{19}{ - \difficulty{5} - - \statement{ - 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. - } -}