Ask Uncle Colin: Scheduling a tournament | Colin

Dear Uncle Colin,

I’m trying to organise a tournament involving seven teams and two pitches. The following conditions must hold:
Each team plays four games
No pair of teams meets more than once
Each team must play at most one pair of back-to-back matches

How would you solve this?
Bit Of Hard Rescheduling

Hi, BOHR, and thanks for your message! I found this quite tricky, but eventually came up with a schedule that works:
Pitch 1
Pitch 2
1 v 7
5 v 3
2 v 1
6 v 4
3 v 2
7 v 5
4 v 3
1 v 6
5 v 4
2 v 7
6 v 5

Continue reading at:


The Mathematical Ninja and the Other Rope | Colin

This is based on a puzzle I heard from @colinthemathmo, who wrote it up here; he heard it from @DavidB52s, and there the trail goes cold.

The Mathematical Ninja lay awake, toes itching. This generally meant that a mission was in the offing. Awake or dreaming? Unclear. But the thought had implanted in the Ninja brain: pull a rope taut around the equator, add six metres to one end and tie it to the other. Now lift the rope upwards until it’s taut. How high do you have to go?

Eyes closed. The rope will follow the equator most of the way around, then take off in a straight line. Let’s call…

Continue reading at:

Ask Uncle Colin: What’s $sqrt{ 100! }$? | Colin

Dear Uncle Colin,

How would I work out $\sqrt(\ln(100!))$ in my head?

– Some Tricks I’d Really Like In Number Games

Hi, STIRLING, and thanks for your message!

I don’t know how you’d do it, but I know how the Mathematical Ninja would!

Stirling’s Approximation1 says that $\ln(n!) \approx n \ln(n) – n + \frac{1}{2}\ln(2\pi n)$.

The Mathematical Ninja would expect you to know that $\ln(100) \approx 4.6$ and that $200\pi \approx 628$, the square root of which is a shade over 25. (25.07 or so, but that’s more accurate than we want here.)

So we have $100 \ln(100) – 100 + \frac{1}{2}…

Continue reading at:

A Textbook Error? | Colin

In class, a student asked to work through a question:

Let $f(x) = \frac{5(x-1)}{(x+1)(x-4)} – \frac{3}{x-4}$.

(a) Show that $f(x)$ can be written as $\frac{2}{x+1}$.

(b)Hence find $f^{-1}(x)$, stating its domain.

The answer they gave was outrageous1.
Part (a)

Part (a) was fine: combine it all into a single fraction as $\frac{5(x-1) – 3(x+1)}{(x+1)(x-4)}$, then simplify the top to get $\frac{2(x-4)}{(x+1)(x-4)} = \frac{2}{x+1}$. Not an enormous challenge for someone with algebraic fraction skills.
First part of (b)

… and the inverse function isn’t much more difficult:…

Continue reading at:

Ask Uncle Colin: It’s Hip To Be Square | Colin

Dear Uncle Colin,

I’m struggling to make any headway with this: find all integers $n$ such that $5 \times 2^n + 1$ is square. Any ideas?

Lousy Expression Being Equalto Square Gives Undue Exasperation

Hi, LEBESGUE, and thanks for your message!

Every mathematician should have a Bag Of Tricks – things they look out for that occasionally make problem-solving easier. Think of it as a mathematical toolkit. For me, it’s things like Clever Regrouping, Fermat’s Little Theorem, modulo arithmetic and – more often than the others – the difference of two squares.

It’s not immediately obvious…

Continue reading at:

Wrong, But Useful: Episode | Colin

In this month’s episode of Wrong, But Useful, we’re joined by @DrSmokyFurby and his handler, Belgin Seymenoglu.

Apologies for the poor audio quality on this call. Dave’s fault, obviously1 .

We discuss:
The Talkdust podcast
(via Adam Atkinson): Life insurance
Superpermutations: new record for n = 7 in the comments on a YouTube post
@pecnut and @mscroggs have a LaTeX package that puts hats on things
Desmos adding support for distributions and all sorts
@mdawesmdawes’s QUIBANS website
Dave has been playing the game The Mind
(Via @peterrowlett), 318,000 combinations of pringles

Continue reading at:

On Epiphanies | Colin

I had a fascinating conversation on Twitter the other day about, I suppose, different modes of solving a problem. Here’s where it started:

Heh. You spend half an hour knee-deep in STEP algebra, solve it, then realise that tweaking the diagram a tiny bit turns it into a two-liner.

— Colin Beveridge (@icecolbeveridge) February 5, 2019

I intended it as a throwaway comment, but it got some interesting responses.

@colinthemathmo (Colin Wright) pointed out:

It’s an interesting conundrum … Not all problems have neat solutions, so we have to…

Continue reading at: