Convert "Partition Products" to typst

src/Warm-Ups/Partition Products/main.tex

@@ -1,57 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: Partition Products}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	Take any positive integer $n$. \par
-	Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... + a_k$ \par
-	Maximize the product $a_1 \times a_2 \times ... \times a_k$
-
-
-
-	\begin{solution}
-
-		\textbf{Interesting Solution:}
-
-		Of course, all $a_i$ should be greater than $1$. \par
-		Also, all $a_i$ should be smaller than four, since $x \leq x(x-2)$ if $x \geq 4$. \par
-		Thus, we're left with sequences that only contain 2 and 3. \par
-		\note{Note that two twos are the same as one four, but we exclude fours for simplicity.}
-
-		\vspace{2mm}
-
-		Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \par
-		The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
-
-		\linehack{}
-
-
-
-		\textbf{Calculus Solution:}
-
-		First, solve this problem for equal, non-integer $a_i$:
-
-		\vspace{2mm}
-
-		We know $n = \prod{a_i}$, thus $\ln(n) = \sum{\ln(a_i)}$. \par
-		If all $a_i$ are equal, we get $\ln(n) = k \times \ln(n / k)$. \par
-		Derive wrt $k$ and set to zero to get $\ln(n / k) = 1$ \par
-		So $k = n / e$ and $n / k = e \approx 2.7$
-
-		\vspace{2mm}
-
-		If we try to approximate this with integers, we get the same solution as above.
-	\end{solution}
-\end{document}

src/Warm-Ups/Partition Products/main.typ

@@ -0,0 +1,41 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Partition Products],
+  by: "Mark",
+)
+
+#problem()
+Take any positive integer $n$. \
+Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ...  a_k$ \
+Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
+
+
+#solution([
+  *Interesting Solution:*
+
+  Of course, all $a_i$ should be greater than $1$. \
+  Also, all $a_i$ should be smaller than four, since $x <= x(x-2)$ if $x >= 4$. \
+  Thus, we're left with sequences that only contain 2 and 3. \
+  #note([Note that two twos are the same as one four, but we exclude fours for simplicity.])
+
+  #v(2mm)
+
+  Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \
+  The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
+
+  #v(8mm)
+
+  *Calculus Solution:*
+
+  First, solve this problem for equal, real $a_i$:
+  #v(2mm)
+  We know $n = product(a_i)$, thus $ln(n) = sum(ln(a_i))$. \
+  If all $a_i$ are equal, we get $ln(n) = k #sym.times ln(n / k)$. \
+  Derive wrt $k$ and set to zero to get $ln(n / k) = 1$ \
+  So $k = n / e$ and $n / k = e #sym.approx 2.7$
+
+  #v(2mm)
+
+  If we try to approximate this with integers, we get the same solution as above.
+])