Added a warm-up

Misc/Warm-Ups/partitions.tex

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+\documentclass[
+	solutions,
+	singlenumbering,
+	nopagenumber
+]{../../resources/ormc_handout}
+\usepackage{../../resources/macros}
+
+
+\title{Warm-Up: Partition Products}
+\subtitle{Prepared by \githref{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}