I’m not normally interested in education stuff, but we’ve had a flurry of emails from various people telling us about their projects, and I’ve got nothing else to do today, so I thought I’d round them up.
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A more equitable statement of the jealous husbands puzzle
Every time I use the jealous husbands river crossing problem, I prefix it with a waffly apology about its formulation. You’ll see what I mean; here’s a standard statement of the puzzle:
Three married couples want to cross a river in a boat that is capable of holding only two people at a time, with the constraint that no woman can be in the presence of another man unless her (jealous) husband is also present. How should they cross the river with the least amount of rowing?
I’m planning to use this again next week. It’s a nice puzzle, good for exercises in problem-solving, particularly for Pólya’s “introduce suitable notation”. I wondered if there could be a better way to formulate the puzzle – one that isn’t so poorly stated in terms of gender equality and sexuality.
Apéryodical: Roger Apéry’s Mathematical Story
This is a guest post by mathematician and maths communicator Ben Sparks.
Roger Apéry: 14th November 1916 – 18th December 1994
100 years ago (on 14th November) was born a Frenchman called Roger Apéry. He died in 1994, is buried in Paris, and upon his tombstone is the cryptic inscription:
\[ 1 + \frac{1}{8} + \frac{1}{27} +\frac{1}{64} + \cdots \neq \frac{p}{q} \]
Apéry’s gravestone – Image from St. Andrews MacTutor Archive
Roger Apéry – Image from St. Andrews MacTutor Archive
The centenary of Roger Apéry’s birth is an appropriate time to unpack something of this mathematical story.
Puzzlebomb – November 2016
Puzzlebomb is a monthly puzzle compendium. Issue 59 of Puzzlebomb, for November 2016, can be found here:
Puzzlebomb – Issue 59 – November 2016 (printer-friendly version)
The solutions to Issue 59 can be found here:
Puzzlebomb – Issue 59 – November 2016 – Solutions
Previous issues of Puzzlebomb, and their solutions, can be found here.
The magic number 25641
Reader of the site Bhaskar Hari Phadke has written in to tell us this fun fact about the number $25641$. It’s easier to show than to describe, so here goes:
\begin{align}
25641 \times \color{blue}{1} \times 4 &= \color{blue}{1}02564 \\
25641 \times \color{blue}{2} \times 4 &= \color{blue}{2}05128 \\
25641 \times \color{blue}{3} \times 4 &= \color{blue}{3}07692 \\
25641 \times \color{blue}{4} \times 4 &= \color{blue}{4}10256 \\
25641 \times \color{blue}{5} \times 4 &= \color{blue}{5}12820 \\
25641 \times \color{blue}{6} \times 4 &= \color{blue}{6}15384 \\
25641 \times \color{blue}{7} \times 4 &= \color{blue}{7}17948 \\
25641 \times \color{blue}{8} \times 4 &= \color{blue}{8}20512 \\
25641 \times \color{blue}{9} \times 4 &= \color{blue}{9}23076
\end{align}
A good one to challenge a young person with.
I did a little bit of Sloanewhacking and found a couple of sequences containing $25641$ which almost, but don’t quite, describe this property. So, semi-spoiler warning: you might enjoy A256005 and A218857. I’d like to come up with the ‘magic number’ which looks the least like it’ll have this property – any ideas?
Thanks, Bhaskar!
Carnival of Mathematics 139
The next issue of the Carnival of Mathematics, rounding up blog posts from the month of September, and compiled by Manjil, is now online at Gonit Sora.
The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.
“π – It’s Complicated” – a talk I gave on Pi Day 2016 at Ustinov College Café Scientifique
I was invited to give a talk for Ustinov College’s Café Scientifique on π Day this year. The turnout wasn’t great and I put quite a bit of effort into the slides, so I wanted to put it online. I’ve finally got hold of the recording, so here it is. Unfortunately they didn’t set the camera’s exposure properly, making the screen illegible, so you’ll probably want to follow along with the slides in another window.
I tried to come up with a way of writing today’s date as a multiple of π Day, but couldn’t make it work. However, I did realise that Halloween (31/10) is the best approximation to π between now and the next π day (I think). Sπooky!