\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}