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 2019-07-03 
This year's Big Internet Math-Off is now underway with 15 completely new contestants (plus one returning contender). As I'm not the returning contestant, I haven't been spending my time preparing my pitches. Instead, I've spent my time making an unofficial Big Internet Math-Off sticker book.
To complete the sticker book, you will need to collect 162 different stickers. Every day, you will be given a pack of 5 stickers; there are also some bonus packs available if you can find them (Hint: keep reading).

How many stickers will I need?

Using the same method as I did for last year's World Cup sticker book, you can work out that the expected number of stickers needed to finish the sticker book:
If you have already stuck \(n\) stickers into your album, then the probability that the next sticker you get is new is
$$\frac{162-n}{162}.$$
The probability that the second sticker you get is the next new sticker is
$$\mathbb{P}(\text{next sticker is not new})\times\mathbb{P}(\text{sticker after next is new})$$ $$=\frac{n}{162}\times\frac{162-n}{162}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next new sticker is
$$\left(\frac{n}{162}\right)^{i-1}\times\frac{162-n}{162}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find a new one:
$$\sum_{i=1}^{\infty}i \left(\frac{162-n}{162}\right) \left(\frac{n}{162}\right)^{i-1} = \frac{162}{162-n}$$
Therefore, to get all 162 stickers, you should expect to buy
$$\sum_{n=0}^{161}\frac{162}{162-n} = 918 \text{ stickers}.$$
Using just your daily packs, it will take you until the end of the year to collect this many stickers. Of course, you'll only need to collect this many if you don't swap your duplicate stickers.

How many stickers will I need if I swap?

To work out the expected number of stickers stickers you'd need if you swap, let's first think about two people who want to complete their stickerbooks together. If there are \(a\) stickers that both collectors need and \(b\) stickers that one collector has and the other one needs, then let \(E_{a,b}\) be the expected number of stickers they need to finish their sticker books. The next sticker they get could be one of three things:
Therefore, the expected number of stickers they need to complete their sticker books is
$$E_{a,b}=1+\frac{a}{162}E_{a-1,b+1}+\frac{b}{162}E_{a,b-1}+\frac{162-a-b}{162}E_{a,b}.$$
This can be rearranged to give
$$E_{a,b}= \frac{162}{a+b}+ \frac{a}{a+b}E_{a-1,b+1} +\frac{b}{a+b}E_{a,b-1} $$
We know that $E_{0,0}=0$ (as if \(a=0\) and \(b=0\), both collectors have already finished their sticker books). Using this and the formula above, we can work out that
$$E_{0,1}=162+E_{0,0}=162$$ $$E_{1,0}=162+E_{0,1}=324$$ $$E_{0,2}=\frac{162}2+E_{0,1}=243$$ $$E_{1,1}=\frac{162}2+\frac12E_{0,2}+\frac12E_{1,0}=364.5$$
... and so on until we find that \(E_{162,0}=1269\), and so our collectors should expect to collect 634 stickers each to complete their sticker books.
For three people, we can work out that if there are \(a\) stickers that all three need, \(b\) stickers that two need, and \(c\) stickers that one needs, then
$$ E_{a,b,c} = \frac{162}{a+b+c}+ \frac{a}{a+b+c}E_{a-1,b+1,c} +\frac{b}{a+b+c}E_{a,b-1,c+1} +\frac{c}{a+b+c}E_{a,b,c-1}. $$
In the same way as for two people, we find that \(E_{162,0,0}=1572\), and so our collectors should expect to collect 524 stickers each to complete their sticker books.
Doing the same thing for four people gives an expected 463 stickers required each.
After four people, however, the Python code I wrote to do these calculations takes too long to run, so instead I approximated the numbers by simulating 500 groups of \(n\) people collecting stickers, and taking the average number of stickers they needed. The results are shown in the graph below.
The red dots are the expected values we calculated exactly, and the blue crosses are the simulated values. It looks like you'll need to collect at least 250 stickers to finish the album: in order to get this many before the end of the Math-Off, you'll need to find 20 bonus packs...
Of course, these are just the mean values and you could get lucky and need fewer stickers. The next graph shows box plots with the quartiles of the data from the simulations.
So if you're lucky, you could complete the album with fewer stickers or fewer friends.
As a thank you for reading to the end of this blog post, here's a link that will give you two bonus packs and help you on your way to the 250 expected stickers...

Similar posts

Runge's Phenomenon
World Cup stickers 2018, pt. 3
Mathsteroids
World Cup stickers 2018, pt. 2

Comments

Comments in green were written by me. Comments in blue were not written by me.
 2019-07-06 
@Pat Ashforth: Thanks, fixed
Reply
Matthew
 2019-07-04 
Link to sticker book, in the first paragraph, does not work. It points to mathoffstickbook.com
Reply
Pat Ashforth
 2019-07-04 
@Road: Thanks, fixed
Reply
Matthew
 2019-07-03 
minor typo for the 2 collector case


> and so our collectors should expect to collect 364 stickers

should be 634.
Reply
Road
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 2019-06-19 
Last night at MathsJam, Peter Kagey showed me a conjecture about OEIS sequence A308092.
A308092
The sum of the first \(n\) terms of the sequence is the concatenation of the first \(n\) bits of the sequence read as binary, with \(a(1) = 1\).
1, 2, 3, 7, 14, 28, 56, 112, 224, 448, 896, 1791, 3583, 7166, ...
To understand this definition, let's look at the first few terms of this sequence written in binary:
1, 10, 11, 111, 1110, 11100, 111000, 1110000, 11100000, 111000000, ...
By "the concatenation of the first \(n\) bits of the sequence", it means the first \(n\) binary digits of the whole sequence written in order: 1, then 11, then 110, then 1101, then 11011, then 110111, and so on. So the definition means:
As we know that the sum of the first \(n-1\) terms is the first \(n-1\) digits, we can calculate the third term of this sequence onwards using: "\(a(n)\) is the concatenation of the first \(n\) bits of the sequence subtract concatenation of the first \(n-1\) bits of the sequence":

The conjecture

Peter's conjecture is that the number of 1s in each term is greater than or equal to the number of 1s in the previous term.
I'm going to prove this conjecture. If you'd like to have a try first, stop reading now and come back when you're ready for spoilers. (If you'd like a hint, read the next section then pause again.)

Adding a digit

The third term of the sequence onwards can be calculated by subtracting the first \(n-1\) digits from the first \(n\) digits. If the first \(n-1\) digits form a binary number \(x\), then the first \(n\) digits will be \(2x+d\), where \(d\) is the \(n\)th digit (because moving all the digits to the left one place in binary is multiplying by two).
Therefore the different is \(2x+d-x=x+d\), and so we can work out the \(n\)th term of the sequence by adding the \(n\)th digit in the sequence to the first \(n-1\) digits. (Hat tip to Martin Harris, who spotted this first.)

Carrying

Adding 1 to a binary number the ends in 1 will cause 1 to carry over to the left. This carrying will continue until the 1 is carried into a position containing 0, and after this all the digits to the left of this 0 will remain unchanged.
Therefore adding a digit to the first \(n-1\) digits can only change the digits from the rightmost 0 onwards.

Endings

We can therefore disregard all the digits before the rightmost 0, and look at how the \(n\)th term compares to the \((n-1)\)th term. There are 5 ways in which the first \(n\) digits could end:
We now look at each of these in turn and show that the \(n\)th term will contain at least as many ones at the \((n-1)\)th term.

Case 1: \(00\)

If the first \(n\) digits of the sequence are \(x00\) (a binary number \(x\) followed by two zeros), then the \((n-1)\)th term of the sequence is \(x+0=x\), and the \(n\)th term of the sequence is \(x0+0=x0\). Both \(x\) and \(x0\) contain the same number of ones.

Case 2: \(010\)

If the first \(n\) digits of the sequence are \(x010\), then the \((n-1)\)th term of the sequence is \(x0+1=x1\), and the \(n\)th term of the sequence is \(x01+0=x01\). Both \(x1\) and \(x01\) contain the same number of ones.

Case 3: \(01...10\)

If the first \(n\) digits of the sequence are \(x01...10\), then the \((n-1)\)th term of the sequence is \(x01...1+1=x10...0\), and the \(n\)th term of the sequence is \(x01...10+1=x01...1\). \(x01...1\) contains more ones than \(x10...0\).

Case 4: \(01\)

If the first \(n\) digits of the sequence are \(x01\), then the \((n-1)\)th term of the sequence is \(x+0=x\), and the \(n\)th term of the sequence is \(x0+1=x1\). \(x1\) contains one more one than \(x\).

Case 5: \(01...1\)

If the first \(n\) digits of the sequence are \(x01...1\), then the \((n-1)\)th term of the sequence is \(x01...1+1=x10...0\), and the \(n\)th term of the sequence is \(x01...1+1=x10...0\). Both these contain the same number of ones.

In all five cases, the \(n\)th term contains more ones or an equal number of ones to the \((n-1)\)th term, and so the conjecture is true.

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 2019-04-09 
In the latest issue of Chalkdust, I wrote an article with Edmund Harriss about the Harriss spiral that appears on the cover of the magazine. To draw a Harriss spiral, start with a rectangle whose side lengths are in the plastic ratio; that is the ratio \(1:\rho\) where \(\rho\) is the real solution of the equation \(x^3=x+1\), approximately 1.3247179.
A plastic rectangle
This rectangle can be split into a square and two rectangles similar to the original rectangle. These smaller rectangles can then be split up in the same manner.
Splitting a plastic rectangle into a square and two plastic rectangles.
Drawing two curves in each square gives the Harriss spiral.
A Harriss spiral
This spiral was inspired by the golden spiral, which is drawn in a rectangle whose side lengths are in the golden ratio of \(1:\phi\), where \(\phi\) is the positive solution of the equation \(x^2=x+1\) (approximately 1.6180339). This rectangle can be split into a square and one similar rectangle. Drawing one arc in each square gives a golden spiral.
A golden spiral

Continuing the pattern

The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.
The rectangles in which golden and Harriss spirals can be drawn.
Continuing the pattern of these arrangements suggests the following rectangle, split into a square and three similar rectangles:
Let the side of the square be 1 unit, and let each rectangle have sides in the ratio \(1:x\). We can then calculate that the lengths of the sides of each rectangle are as shown in the following diagram.
The side lengths of the large rectangle are \(\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1\) and \(\frac1{x^2}+\frac1x+1\). We want these to also be in the ratio \(1:x\). Therefore the following equation must hold:
$$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1=x\left(\frac1{x^2}+\frac1x+1\right)$$
Rearranging this gives:
$$x^4-x^2-x-1=0$$ $$(x+1)(x^3-x^2-1)=0$$
This has one positive real solution:
$$x=\frac13\left( 1 +\sqrt[3]{\tfrac12(29-3\sqrt{93})} +\sqrt[3]{\tfrac12(29+3\sqrt{93})} \right).$$
This is equal to 1.4655712... Drawing three arcs in each square allows us to make a spiral from a rectangle with sides in this ratio:
A spiral which may or may not have a name yet.

Continuing the pattern

Adding a fourth rectangle leads to the following rectangle.
The side lengths of the largest rectangle are \(1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4}\) and \(1+\frac2x+\frac1{x^2}+\frac1{x^3}\). Looking for the largest rectangle to also be in the ratio \(1:x\) leads to the equation:
$$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4} = x\left(1+\frac2x+\frac1{x^2}+\frac1{x^3}\right)$$ $$x^5+x^4-x^3-2x^2-x-1 = 0$$
This has one real solution, 1.3910491... Although for this rectangle, it's not obvious which arcs to draw to make a spiral (or maybe not possible to do it at all). But at least you get a pretty fractal:

Continuing the pattern

We could, of course, continue the pattern by repeatedly adding more rectangles. If we do this, we get the following polynomials and solutions:
Number of rectanglesPolynomialSolution
1\(x^2 - x - 1=0\)1.618033988749895
2\(x^3 - x - 1=0\)1.324717957244746
3\(x^4 - x^2 - x - 1=0\)1.465571231876768
4\(x^5 + x^4 - x^3 - 2x^2 - x - 1=0\)1.391049107172349
5\(x^6 + x^5 - 2x^3 - 3x^2 - x - 1=0\)1.426608021669601
6\(x^7 + 2x^6 - 2x^4 - 3x^3 - 4x^2 - x - 1=0\)1.4082770325090774
7\(x^8 + 2x^7 + 2x^6 - 2x^5 - 5x^4 - 4x^3 - 5x^2 - x - 1=0\)1.4172584399350432
8\(x^9 + 3x^8 + 2x^7 - 5x^5 - 9x^4 - 5x^3 - 6x^2 - x - 1=0\)1.412713760332943
9\(x^{10} + 3x^9 + 5x^8 - 5x^6 - 9x^5 - 14x^4 - 6x^3 - 7x^2 - x - 1=0\)1.414969877544769
The numbers in this table appear to be heading towards around 1.414, or \(\sqrt2\). This shouldn't come as too much of a surprise because \(1:\sqrt2\) is the ratio of the sides of A\(n\) paper (for \(n=0,1,2,...\)). A0 paper can be split up like this:
Splitting up a piece of A0 paper
This is a way of splitting up a \(1:\sqrt{2}\) rectangle into an infinite number of similar rectangles, arranged following the pattern, so it makes sense that the ratios converge to this.

Other patterns

In this post, we've only looked at splitting up rectangles into squares and similar rectangles following a particular pattern. Thinking about other arrangements leads to the following question:
Given two real numbers \(a\) and \(b\), when is it possible to split an \(a:b\) rectangle into squares and \(a:b\) rectangles?
If I get anywhere with this question, I'll post it here. Feel free to post your ideas in the comments below.

Similar posts

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Comments

Comments in green were written by me. Comments in blue were not written by me.
 2019-05-02 
@g0mrb: CORRECTION: There seems to be no way to correct the glaring error in that comment. A senior moment enabled me to reverse the nomenclature for paper sizes. Please read the suffixes as (n+1), (n+2), etc.
Reply
(anonymous)
 2019-05-02 
I shall remain happy in the knowledge that you have shown graphically how an A(n) sheet, which is 2 x A(n-1) rectangles, is also equal to the infinite series : A(n-1) + A(n-2) + A(n-3) + A(n-4) + ... Thank-you, and best wishes for your search for the answer to your question.
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g0mrb
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 2019-03-26 
I originally wrote this post for The Aperiodical.
A few months ago, Adam Townsend went to lunch and had a conversation. I wasn't there, but I imagine the conversation went something like this:
Adam: Hello.
Smitha: Hello.
Adam: How are you?
Smitha: Not bad. I've had a funny idea, actually.
Adam: Yes?
Smitha: You know how the \hat command in LaTeΧ puts a caret above a letter?... Well I was thinking it would be funny if someone made a package that made the \hat command put a picture of an actual hat on the symbol instead?
Adam: (After a few hours of laughter.) I'll see what my flatmate is up to this weekend...
Jeff: What on Earth are you two talking about?!
As anyone who has been anywhere near maths at a university in the last ∞ years will be able to tell you, LaTeΧ is a piece of maths typesetting software. It's a bit like a version of Word that runs in terminal and makes PDFs with really pretty equations.
By default, LaTeΧ can't do very much, but features can easily added by importing packages: importing the graphicsx package allows you to put images in your PDF; importing geometry allows you to easily change the page margins; and importing realhats makes the \hat command put real hats above symbols.

Changing the behaviour of \hat

By default, the LaTeΧ command \hat puts a pointy "hat" above a symbol:
a (left) and \hat{a} (right)
After Adam's conversation, we had a go at redefining the \hat command by putting the following at the top of our LaTeΧ file.
 LaTeΧ 
\renewcommand{\hat}[1]{
    % We put our new definition here
}

After a fair amount of fiddling with the code, we eventually got it to produce the following result:
a (left) and \hat{a} (right) while using the realhats package
We were now ready to put our code into a package so others could use it.

How to write a package

A LaTeΧ package is made up of:
It's quite common to make the first two of these by making a dtx file and an ins file. And no, we have no idea either why these are the file extensions used or why this is supposedly simpler than making a sty file and a PDF.
The ins file says which bits of the dtx should be used to make up the sty file. Our ins file looks like this:
 LaTeΧ 
\input{docstrip.tex}
\keepsilent
\usedir{tex/latex/realhats}
\preamble
 *License goes here*
\endpreamble
\askforoverwritefalse
\generate{
  \file{realhats.sty}{\from{realhats.dtx}{realhats}}
}
\endbatchfile
The most important command in this file is \generate: this says that that the file realhats.sty should be made from the file realhats.dtx taking all the lines that are marked as part of realhats. The following is part of our dtx file:
 LaTeΧ 
%\lstinline{realhats} is a package for \LaTeX{} that makes the \lstinline{\hat}
%command put real hats on symbols.
%For example, the input \lstinline@\hat{a}=\hat{b}@ will produce the output:
%\[\hat{a}=\hat{b}\]
%To make a vector with a hat, the input \lstinline@\hat{\mathbf{a}}@ produces:
%\[\hat{\mathbf{a}}\]
%
%\iffalse
%<*documentation>
\documentclass{article}
\usepackage{realhats}
\usepackage{doc}
\usepackage{listings}
\title{realhats}
\author{Matthew W.~Scroggs \& Adam K.~Townsend}
\begin{document}
\maketitle
    \DocInput{realhats.dtx}
\end{document}
%</documentation>
%\fi

%\iffalse
%<*realhats>
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{realhats}[2019/02/02 realhats]

\RequirePackage{amsmath}
\RequirePackage{graphicx}
\RequirePackage{ifthen}
\renewcommand{\hat}[1]{
    % We put our new definition here
}
%</realhats>
%\fi
The lines near the end between <*realhats> and </realhats> will be included in the sty file, as they are marked at part of realhats.
The rest of this file will make the PDF documentation when the dtx file is compiled. The command \DocInput tells LaTeΧ to include the dtx again, but with the %s that make lines into comments removed. In this way all the comments that describe the functionality will end up in the PDF. The lines that define the package will not be included in the PDF as they are between \iffalse and \fi.
Writing both the commands and the documentation in the same file like this means that the resulting file is quite a mess, and really quite ugly. But this is apparently the standard way of writing LaTeΧ packages, so rest assured that it's not just our code that ugly and confusing.

What to do with your package

Once you've written a package, you'll want to get it out there for other people to use. After all, what's the point of being able to put real hats on top of symbols if the whole world can't do the same?
First, we put the source code of our package on GitHub, so that Adam and I had an easy way to both work on the same code. This also allows other LaTeΧ lovers to see the source and contribute to it, although none have chosen to add anything yet.
Next, we submitted our package to CTAN, the Comprehensive TeΧ Archive Network. CTAN is an archive of thousands of LaTeΧ packages, and putting realhats there gives LaTeΧ users everywhere easy access to real hats. Within days of being added to CTAN, realhats was added (with no work by us) to MikTeX and TeX Live to allow anyone using these LaTeΧ distributions to seemlessly install it as soon as it is needed.
We figured that the packaged needed a website too, so we made one. We also figured that the website should look as horrid as possible.

How to use realhats

So if you want to end fake hats and put real hats on top of your symbols, you can simply write \usepackage{realhats} at the top of your LaTeΧ file.

Similar posts

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 2019-01-01 
It's 2019, and the Advent calendar has disappeared, so it's time to reveal the answers and annouce the winners. But first, some good news: with your help, Santa was able to work out who had stolen the presents and save Christmas:
Now that the competition is over, the questions and all the answers can be found here. Before announcing the winners, I'm going to go through some of my favourite puzzles from the calendar, reveal the solution and a couple of notes and Easter eggs.

Highlights

My first highlight is the first puzzle in the calendar. This is one of my favourites as it has a pleasingly neat solution involving a surprise appearance of a very famous sequence.

1 December

There are 5 ways to write 4 as the sum of 1s and 2s:
  • 1+1+1+1
  • 2+1+1
  • 1+2+1
  • 1+1+2
  • 2+2
Today's number is the number of ways you can write 12 as the sum of 1s and 2s.

Show answer


My next highlight is a puzzle that I was particularly proud of cooking up: again, this puzzle at first glance seems like it'll take a lot of brute force to solve, but has a surprisingly neat solution.

10 December

The equation \(x^2+1512x+414720=0\) has two integer solutions.
Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.

Show answer


My next highlight is a geometry problem that appears to be about polygons, but actually it's about a secret circle.

12 December

These three vertices form a right angled triangle.
There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometime the three vertices you pick form a right angled triangle.
Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.

 

Show answer


My final highlight is a puzzle about the expansion of a fraction in different bases.

22 December

In base 2, 1/24 is 0.0000101010101010101010101010...
In base 3, 1/24 is 0.0010101010101010101010101010...
In base 4, 1/24 is 0.0022222222222222222222222222...
In base 5, 1/24 is 0.0101010101010101010101010101...
In base 6, 1/24 is 0.013.
Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.
Today's number is the smallest base in which 1/10890 has a finite number of digits.
Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).

Show answer

Notes and Easter eggs

I had a lot of fun this year coming up with the names for the possible theives. In order to sensibly colour code each suspect's clues, there is a name of a colour hidden within each name: Fred Metcalfe, Jo Ranger, Bob Luey, Meg Reeny, and Kip Urples. Fred Metcalfe's colour is contained entirely within his forename, so you may be wondering where his surname came from. His surname is shared with Paul Metcalfe—the real name of a captain whose codename was a certain shade of red.
On 20 December, Elijah Kuhn emailed me to point out that it was possible to solve the final puzzle a few days early: although he could not yet work out the full details of everyone's timetable, he had enough information to correctly work out who the culprit was and between which times the theft had taken place.
Once you've entered 24 answers, the calendar checks these and tells you how many are correct. This year, I logged the answers that were sent for checking and have looked at these to see which puzzles were the most and least commonly incorrect. The bar chart below shows the total number of incorrect attempts at each question.
You can see that the most difficult puzzles were those on 13, 24, and 10 December; and the easiest puzzles were on 5, 23, 11, and 15 December.
I also snuck a small Easter egg into the door arrangement: the doors were arranged to make a magic square, with each row and column, plus the two diagonals, adding to 55.

The solution

The solutions to all the individual puzzles can be found here. Using the clues, you can work out that everyone's seven activities formed the following timetable.
Bob LueyFred MetcalfeJo RangerKip UrplesMeg Reeny
0:00–1:21
Billiards
0:00–2:52
Maths puzzles
0:00–2:33
Maths puzzles
0:00–1:21
Billiards
0:00–1:10
Ice skating
1:10–2:33
Skiing
1:21–2:52
Ice skating
1:21–2:52
Stealing presents
2:33–4:45
Billiards
2:33–4:45
Billiards
2:52–3:30
Lunch
2:52–3:30
Lunch
2:52–3:30
Lunch
3:30–4:45
Climbing
3:30–4:45
Climbing
3:30–4:45
Climbing
4:45–5:42
Curling
4:45–5:42
Curling
4:45–5:42
Curling
4:45–5:42
Curling
4:45–5:42
Lunch
5:42–7:30
Maths puzzles
5:42–7:30
Ice skating
5:42–7:30
Chess
5:42–7:30
Chess
5:42–7:30
Maths puzzles
7:30–10:00
Skiing
7:30–9:45
Chess
7:30–8:45
Skiing
7:30–10:00
Maths puzzles
7:30–9:45
Chess
8:45–9:45
Lunch
9:45–10:00
Table tennis
9:45–10:00
Table tennis
9:45–10:00
Table tennis
Following your investigation, Santa found all the presents hidden under Kip Urples's bed, fired Kip and sucessfully delivered all the presents on Christmas Eve.

The winners

And finally (and maybe most importantly), on to the winners: 73 people submitted answers to the final logic puzzle. Their (very) approximate locations are shown on this map:
From the correct answers, the following 10 winners were selected:
1Sarah Brook
2Mihai Zsisku
3Bhavik Mehta
4Peter Byrne
5Martin Harris
6Gert-Jan de Vries
7Lyra
8James O'Driscoll
9Harry Poole
10Albert Wood
Congratulations! Your prizes will be on their way shortly. Additionally, well done to Alan Buck, Alex Ayres, Alex Bolton, Alex Lam, Alexander Ignatenkov, Alexandra Seceleanu, Andrew Turner, Ashwin Agarwal, Becky Russell, Ben Reiniger, Brennan Dolson, Carl Westerlund, Cheng Wai Koo, Christopher Embrey, Corbin Groothuis, Dan Whitman, David, David Ault, David Kendel, Dennis Oltmanns, Elijah Kuhn, Eric, Eric Kolbusz, Evan Louis Robinson, Felix Breton, Fred Verheul, Gregory Loges, Hannah, Jean-Noël Monette, Jessica Marsh, Joe Gage, Jon Palin, Jonathan Winfield, Kai Lam, Louis de Mendonca, M Oostrom, Martine Vijn Nome, Matt Hutton, Matthew S, Matthew Wales, Michael DeLyser, MikeKim, Naomi Bowler, Pranshu Gaba, Rachel Bentley, Raymond Arndorfer, Rick Simineo, Roni, Rosie Paterson, Sam Hartburn, Scott, Sheby, Shivanshi, Stephen Cappella, Steve Paget, Thomas Smith, Tony Mann, Valentin Vălciu, Yasha Ayyari, Zack Wolske, and Zoe Griffiths, who all also submitted the correct answer but were too unlucky to win prizes this time.
See you all next December, when the Advent calendar will return.

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