mirror of
https://github.com/Rushilwiz/math381.git
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269 lines
7.7 KiB
TeX
269 lines
7.7 KiB
TeX
\documentclass{article}
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\usepackage{fancyhdr}
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\usepackage{extramarks}
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\usepackage{amsmath}
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\usepackage{amsthm}
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\usepackage{amsfonts}
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\usepackage{tikz}
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\usepackage[plain]{algorithm}
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\usepackage{algpseudocode}
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\usepackage[shortlabels]{enumitem}
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\usepackage{mathtools}
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\usepackage{amssymb}
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\usetikzlibrary{automata,positioning}
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%
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% Basic Document Settings
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%
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\topmargin=-0.45in
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\evensidemargin=0in
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\oddsidemargin=0in
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\textwidth=6.5in
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\textheight=9.0in
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\headsep=0.25in
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\linespread{1.1}
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\pagestyle{fancy}
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\lhead{\hmwkAuthorName}
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\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}
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\lfoot{\lastxmark}
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\cfoot{\thepage}
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\renewcommand\headrulewidth{0.4pt}
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\renewcommand\footrulewidth{0.4pt}
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\setlength\parindent{0pt}
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%
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% Create Problem Sections
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%
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\newcommand{\enterProblemHeader}[1]{
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\nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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}
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\newcommand{\exitProblemHeader}[1]{
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\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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\stepcounter{#1}
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\nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
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}
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\setcounter{secnumdepth}{0}
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\newcounter{partCounter}
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\newcounter{homeworkProblemCounter}
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\setcounter{homeworkProblemCounter}{1}
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\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}
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\newcommand{\hmwkTitle}{Homework 8}
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\newcommand{\hmwkDueDate}{November 16, 2023}
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\newcommand{\hmwkClass}{Discrete Math}
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\newcommand{\hmwkClassTime}{Section 003}
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\newcommand{\hmwkClassInstructor}{Reese Lance}
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\newcommand{\hmwkAuthorName}{\textbf{Rushil Umaretiya}}
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%
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% Title Page
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%
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\title{
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\vspace{2in}
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\textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
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\normalsize\vspace{0.1in}\small{\textbf{Due\ on\ \hmwkDueDate\ at 11:59pm}}\\
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\normalsize\text{Tuesday/Thursday 11:00-12:15, Phillips 383}\\
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\vspace{0.1in}\large{\textit{\hmwkClassInstructor\ - \hmwkClassTime}}
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\vspace{3in}
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}
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\author{\hmwkAuthorName\\\small{rumareti@unc.edu}}
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\date{}
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\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}
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%
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% Various Helper Commands
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%
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% Useful for algorithms
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\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
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% For derivatives
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\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
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% For partial derivatives
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\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
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% Integral dx
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\newcommand{\dx}{\mathrm{d}x}
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% Alias for the Solution section header
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\newcommand{\solution}{\textbf{\large Solution}}
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\newcommand{\unit}[1]{\section{Unit #1}}
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\newcommand{\problem}[1]{\textbf{\##1}}
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\newcommand{\prob}[1]{\problem{#1}}
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% Probability commands: Expectation, Variance, Covariance, Bias
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\newcommand{\E}{\mathrm{E}}
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\newcommand{\Var}{\mathrm{Var}}
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\newcommand{\Cov}{\mathrm{Cov}}
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\newcommand{\Bias}{\mathrm{Bias}}
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\renewcommand{\And}{\wedge}
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\newcommand{\Or}{\vee}
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\newcommand{\Xor}{\oplus}
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\newcommand{\Not}{\neg}
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\newcommand{\Implies}{\rightarrow}
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\newcommand{\Iff}{\leftrightarrow}
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\newcommand{\union}{\cup}
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\newcommand{\intersection}{\cap}
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\newcommand{\AllIntegers}{\mathbb{Z}}
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\newcommand{\AllNaturals}{\mathbb{N}}
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\newcommand{\AllRationals}{\mathbb{Q}}
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\newcommand{\AllReals}{\mathbb{R}}
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\newcommand{\AllComplexes}{\mathbb{C}}
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\renewcommand{\mod}{\textbf{ mod }}
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\renewcommand{\pmod}[1]{\textbf{ (mod }#1)}
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\begin{document}
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\maketitle
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\pagebreak
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\unit{4.1}
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\prob{10}
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Prove that if \(a\) and \(b\) are nonzero integers, \(a | b\), and \(a + b\) is odd, then \(a\) is odd.
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\begin{proof}
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Direct proof. Assume \(a\) and \(b\) are nonzero integers, \(a | b\), and \(a + b\) is odd. Then, by definition of divisibility, \(\exists k \in \AllIntegers, ak = b\). We also know that \(a + b\) is odd, so \(\exists j \in \AllIntegers, 2j + 1 = a + b\). Substituting \(b\) for \(ak\), we get \(2j + 1 = a + ak\). Factoring out \(a\), we get \(2j + 1 = a(1 + k)\). Since \(1 + k\) is an integer, we know that \(a\) is odd.
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\end{proof}
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\pagebreak
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\prob{26}
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Evaluate these quantites:
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\begin{enumerate}[a)]
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\item \(-17 \textbf{ mod } 2\) = \hspace{12pt} 1
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\item \(144 \textbf{ mod } 7\) = \hspace{14pt} 4
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\item \(-101 \textbf{ mod } 13\) = \hspace{5pt}3
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\item \(199 \textbf{ mod } 19\) = \hspace{9pt} 9
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\end{enumerate}
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\pagebreak
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\prob{34}
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Decide whether each of these integers is congruent to 3 modulo 7.
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\begin{enumerate}[a)]
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\item 37
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\begin{align*}
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7 \nmid (37 - 3)\\
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37 \not\equiv 3 \pmod{7}
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\end{align*}
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\item 66
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\begin{align*}
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7 \mid (66 - 3)\\
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66 \equiv 3 \pmod{7}
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\end{align*}
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\item -17
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\begin{align*}
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7 \nmid (-17 - 3)\\
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-17 \not\equiv 3 \pmod{7}
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\end{align*}
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\item -67
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\begin{align*}
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7 \mid (-67 - 3)\\
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-67 \equiv 3 \pmod{7}
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\end{align*}
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\end{enumerate}
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\pagebreak
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\prob{36}
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Find each of these values.
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\textbf{Note: } The following rules exist for modular arithmetic where \(a, b, n \in \AllIntegers, n \geq 2\):
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\begin{align*}
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(a + b) \textbf{ mod } n &= ((a \textbf{ mod } n) + (b \textbf{ mod } n)) \textbf{ mod } n & \text{Addition Law}\\
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(a \cdot b) \textbf{ mod } n &= ((a \textbf{ mod } n) \cdot (b \textbf{ mod } n)) \textbf{ mod } n & \text{Multiplication Law}\\
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\end{align*}
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\begin{enumerate}[a)]
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\item \((177 \mod 31 + 270 \mod 31) \mod 31\)
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\begin{align*}
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(177 \mod 31 + 270 \mod 31) \mod 31 &= (177 + 270) \mod 31\\
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&= 447 \mod 31\\
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&= 13
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\end{align*}
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\item \((177 \mod 31 \cdot 270 \mod 31) \mod 31\)
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\begin{align*}
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(177 \mod 31 \cdot 270 \mod 31) \mod 31 &= (177 \cdot 270) \mod 31\\
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&= 47790 \mod 31\\
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&= 19
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\end{align*}
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\end{enumerate}
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\pagebreak
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\unit{6.1}
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\prob{8}
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How many different three-letter initials with none of the letters repeated can people have?
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\begin{align*}
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26 \cdot 25 \cdot 24 = 15600
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\end{align*}
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\pagebreak
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\prob{22}
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How many positive integers less than 1000,
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\begin{enumerate}[a)]
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\item are divisible by 7?
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\begin{align*}
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\lfloor \frac{1000}{7} \rfloor = 142
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\end{align*}
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\item are divisible by 7 but not by 11?
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\begin{align*}
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\lfloor \frac{1000}{7} \rfloor - \lfloor \frac{1000}{77} \rfloor = 142 - 12 = 130
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\end{align*}
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\item are divisible by both 7 and 11?
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\begin{align*}
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\lfloor \frac{1000}{77} \rfloor = 12
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\end{align*}
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\item are divisible by either 7 or 11?
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\begin{align*}
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\lfloor \frac{1000}{7} \rfloor + \lfloor \frac{1000}{11} \rfloor - \lfloor \frac{1000}{77} \rfloor = 142 + 90 - 12 = 220
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\end{align*}
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\item are divisible by exactly one of 7 and 11?
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\begin{align*}
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\lfloor \frac{1000}{7} \rfloor + \lfloor \frac{1000}{11} \rfloor - 2 \lfloor \frac{1000}{77} \rfloor = 142 + 90 - 24 = 208
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\end{align*}
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\item are divisible by neither 7 nor 11?
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\begin{align}
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999 - 220 = 779
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\end{align}
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\item have distinct digits?
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\begin{align*}
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9 && \text{One digit}\\
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(10-1) \cdot 9 && \text{Two digits}\\
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(10-1) \cdot 9 \cdot 8 && \text{Three digits}\\
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9 + 9 \cdot 9 + 9 \cdot 8 \cdot 7 = 738 && \text{Total}
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\end{align*}
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\item have distinct digits and are even
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\begin{align*}
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4 && \text{One digit}\\
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9 + 8 \cdot 4 && \text{Two digits}\\
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9 \cdot 8 + 8 \cdot 8 \cdot 4 && \text{Three digits}\\
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4 + (9 + 8 \cdot 4) + (9 \cdot 8 + 8 \cdot 8 \cdot 4) = 373 && \text{Total}
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\end{align*}
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\end{enumerate}
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\pagebreak
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\prob{28}
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How many license plates can be made using either three digits followed by three uppercase English letters or three uppercase English letters followed by three digits?
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\begin{align*}
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10^3 \cdot 26^3 + 26^3 \cdot 10^3 = 3.5152 \cdot 10^7
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\end{align*}
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\pagebreak
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\end{document} |