handouts/Misc/Warm-Ups/partitions.tex
2024-04-01 21:50:50 -07:00

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\documentclass[
solutions,
singlenumbering,
nopagenumber
]{../../resources/ormc_handout}
\usepackage{../../resources/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}