<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[../main.tex]{subfiles}

\begin{document}
	\section{The Euclidean Algorithm}

	\definition{}
	The \textit{greatest common divisor} of $a$ and $b$ is the greatest integer that divides both $a$ and $b$. \\
	We denote this number with $\gcd(a, b)$. For example, $\gcd(45, 60) = 15$.


	\theorem{The Division Algorithm}
	Given two integers $a, b$, we can find two integers $q, r$, where $0 \leq r &lt; b$ and $a = qb + r$. \\
	In other words, we can divide $a$ by $b$ to get $q$ remainder $r$.

	\theorem{}&lt;gcd_abc&gt;
	For any integers $a, b, c$, \\
	$\gcd(ac + b, a) = \gcd(a, b)$

	\problem{}
	Find $\gcd(20, 14)$ by hand.

	\begin{solution}
		$\gcd(20, 14) = 2$
	\end{solution}

	\vfill

	\problem{}&lt;euclid_algorithm&gt;
	Using the theorems above, detail an algorithm for finding $\gcd(a, b)$.\\
	Then, compute $\gcd(1610, 207)$ by hand. \\
	Have an instructor check your work before moving on.

	\begin{solution}
		Using \ref{gcd_abc} and the division algorthm,

		% Minipage prevents column breaks inside body
		\begin{multicols}{2}
			\begin{minipage}{\columnwidth}
				$\gcd(1610, 207)$ \\
				$= \gcd(207, 161)$ \\
				$= \gcd(161, 46)$ \\
				$= \gcd(46, 23)$ \\
				$= \gcd(23, 0) = 23$ \\
			\end{minipage}

			\columnbreak

			\begin{minipage}{\columnwidth}
				$1610 = 207 \times 7 + 161$ \\
				$207 = 161 \times 1 + 46$ \\
				$161 = 46 \times 3 + 23$ \\
				$46 = 23 \times 2 + 0$ \\
			\end{minipage}
		\end{multicols}
	\end{solution}

	\vfill
	\pagebreak

	\problem{Divide and Conquer}
	If we are given $a, b, c$, when can we find $u, v$ that satisfy $au + bv = c$?

	\problempart{Divide}
	Show that if we find a solution $(u, v)$ to $au + bv = \gcd(a, b)$, we can easily find a $(u, v)$ for any other value of $c$. \\
	\textcolor{gray}{\textit{Note: } We are not looking for \textit{all} $(u, v)$ that solve $au + bv = c$, we are looking for an easy way to find \textit{any} $(u, v)$.}

	\begin{solution}
		Note that $\gcd(a, b)$ divides both a and b. \\
		Therefore, any $c$ must be divisible by $\gcd(a, b)$.
		The smallest such $c$ is $\gcd(a, b)$ itself, and we can get all other tuples $(u, v, c)$ by scaling.
	\end{solution}

	\vfill

	\problempart{Conquer}&lt;extend_e_algorithm&gt;
	Using the output of your algorithm\footnotemark{} from \ref{euclid_algorithm},
	\footnotetext{Your solution to \ref{euclid_algorithm} is called the \textit{Euclidean Algorithm}}
	\begin{itemize}
		\item[-] find a pair $(u, v)$ that satisfies $20u + 14v = \gcd(20, 14)$
		\item[-] find a pair $(u, v)$ that satisfies $541u + 34v = \gcd(541, 34)$ \\
		% gcd = 1
		% u = 11; v = -175
	\end{itemize}
	For which numbers $c$ can we find a $(u, v)$ so that $541u + 34v = c$? \\
	For every such $c$, what are $u$ and $v$?

	\begin{solution}

		Using the output of the Euclidean Algorithm, we can use substitution and a bit of algebra to solve such problems. Consider the following example:

		\begin{multicols}{2}
			\begin{minipage}{\columnwidth}
				\textit{Euclidean Algorithm:} \\
				$20 = 14 \times 1 + 6$ \\
				$14 = 6 \times 2 + 2$ \\
				$6 = 2 \times 3 + 0$ \\
			\end{minipage}

			\columnbreak

			\begin{minipage}{\columnwidth}
				\textit{Rearranged:} \\
				$6 = 20 - 14 \times 1$ \\
				$2 = 14 - 6 \times 2 = \gcd(20, 14)$ \\
			\end{minipage}
		\end{multicols}

		Using the right table, we can replace $6$ in $2 = 14 - 6 \times 2$ to get
		$2 = 14 - (20 - 14) \times 2$, \\
		which gives us $2 = \gcd(20, 14) = (3)14 + (-2)20$. \\

		\textcolor{gray}{\textit{Note to instructors:} You can present the $(20, 14)$ case as an example.}

		\linehack{}

		$(-2)20 + (3)14 = \gcd(20, 14) = 2$ \\
		$(11)541 + (-175)34 = \gcd(541, 34) = 1$

		\linehack{}

		We can find a solution $(u, v)$ when $c$ is any integer multiple of $\gcd(541, 34)$. \\
		If $c = k \times \gcd(541, 34)$, \\
		$u = k \times u_0 = 11k$ and $v = k \times v_0 = -175k$. \\
		(See Part A)

	\end{solution}
	\vfill
	\pagebreak
\end{document}
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