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2018-12-08
Just like last year and the year before, TD and I spent some time in November this year designing a Chalkdust puzzle Christmas card.
![](javascript:showlimage("card2018.jpg"))
The card looks boring at first glance, but contains 10 puzzles. By splitting the answers into pairs of digits, then drawing lines between the dots on the cover for each pair of digits (eg if an answer is 201304, draw a line from dot 20 to dot 13 and another line from dot 13 to dot 4), you will reveal a Christmas themed picture. Colouring the region of the card labelled R red or orange will make this picture even nicer.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically be split into pairs of digits, lines will be drawn between the pairs, and the red region will be coloured...
If you enjoy these puzzles, then you'll almost certainly enjoy this year's puzzle Advent calendar.
| 1. | What is the smallest four digit number whose digits add up to 6? | Answer |
| 2. | What is the length of the hypotenuse of a right angled triangle whose two shorter sides have lengths 152,560 and 114,420? | Answer |
| 3. | What is the lowest common multiple of 1346 and 196? | Answer |
| 4. | What is the sum of all the odd numbers between 0 and 698? | Answer |
| 5. | How many numbers are there between 100 and 10,000 that contain no 0, 1, 2, or 3? | Answer |
| 6. | How many factors (including 1 and the number itself) does the number \(2^{13}\times3^{19}\times5^9\times7^{39}\) have? | Answer |
| 7. | In a book with pages numbered from 1 to 16,020,308, what do the two page numbers on the centre spread add up to? | Answer |
| 8. | You think of a number, then make a second number by removing one of its digits. The sum of these two numbers is 18,745,225. What was your first number? | Answer |
| 9. | What is the largest number that cannot be written as \(13a+119b\), where \(a\) and \(b\) are positive integers or 0? | Answer |
| 10. | You start at the point (0,0) and are allowed to move one unit up or one unit right. How many different paths can you take to get to the point (7,6)? | Answer |
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⭐ top comment (2018-12-09) ⭐
Someone told me I would like this puzzle and they were right!Blaine
×1 @Carmel: It's not meant to check your answers. It only shows up red if the number you enter cannot be split into valid pairs (eg the number has an odd number of digits or one of the pairs of digits is greater than 20).
Matthew
Great puzzle problems! Hint on #9: try starting with an analogous problem using smaller numbers (e.g. 3a + 10b). This helped me to see what I had to do more generally. ×1
Noah
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2018-11-25
This year, the front page of mscroggs.co.uk will once again feature an advent calendar, just like last year, the year before and the year before. Behind each door, there will be a puzzle with a three digit solution. The solution to each day's puzzle forms part of a logic puzzle:
It's nearly Christmas and something terrible has happened: one of Santa's five helpers—Fred Metcalfe, Bob Luey, Meg Reeny, Jo Ranger, and Kip Urples—has stolen all the presents during the North Pole's annual Sevenstival. You need to find the culprit before Christmas is ruined for everyone.
Every year in late November, Santa is called away from the North Pole for a ten hour meeting in which a judgemental group of elders decide who has been good and who has been naughty. While Santa is away, it is traditional for his helpers celebrate Sevenstival.
Sevenstival gets its name from the requirement that every helper must take part in exactly seven activities during the celebration; this year's
available activities were billiards, curling, having lunch, solving maths puzzles, table tennis, skiing, chess, climbing and ice skating.
Each activity must be completed in one solid block: it is forbidden to spend some time doing an activity, take a break to do something else then return to the first activity.
This year's Sevenstival took place between 0:00 and 10:00 (North Pole standard time).
During this year's Sevenstival, one of Santa's helpers spent the time for one of their seven activities stealing all the presents from Santa's workshop.
Santa's helpers have 24 pieces of information to give to you, but the culprit is going to lie about everything in an attempt to confuse you, so be careful who you trust.
Behind each day (except Christmas Day), there is a puzzle with a three-digit answer.
Each of these answers forms part of a fact that one of the helpers tells you.
You must work out who the culprit is and between which times the theft took place.
Ten randomly selected people who solve all the puzzles and submit their answers to the logic puzzle using the form behind the door on the 25th will win prizes!
![](javascript:showlimage("prizes2018.jpg"))
The winners will also receive one of these medals:
![](javascript:showlimage("medal-2018.jpg"))
As you solve the puzzles, your answers will be stored. To share your stored answers between multiple devices, enter your email address below the calendar and you will be emailed a magic link to visit on your other devices.
Behind the door on Christmas Day, there will be a form allowing you to submit your answers. The winner will be randomly chosen from all those who submit the correct answer before the end of 2018. Each day's puzzle (and the entry form on Christmas Day) will be available from 5:00am GMT. But as the winners will be selected randomly, there's no need to get up at 5am on Christmas Day to enter!
To win a prize, you must submit your entry before the end of 2018. Only one entry will be accepted per person. If you have any questions, ask them in the comments below or on Twitter.
So once December is here, get solving! Good luck and have a very merry Christmas!
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2018-09-13
This is a post I wrote for round 2 of The Aperiodical's Big Internet Math-Off 2018. As I went out in round 1 of the Big Math-Off, you got to read about the real projective plane instead of this.
Polynomials are very nice functions: they're easy to integrate and differentiate, it's quick to calculate their value at points, and they're generally friendly to deal with. Because of this, it can often be useful to find a polynomial that closely approximates a more complicated function.
Imagine a function defined for \(x\) between -1 and 1. Pick \(n-1\) points that lie on the function. There is a unique degree \(n\) polynomial (a polynomial whose highest power of \(x\) is \(x^n\)) that passes through these points. This polynomial is called an interpolating polynomial, and it sounds like it ought to be a pretty good approximation of the function.
So let's try taking points on a function at equally spaced values of \(x\), and try to approximate the function:
$$f(x)=\frac1{1+25x^2}$$
![](javascript:showlimage("runge-uniform.gif"))
Polynomial interpolations of \(\displaystyle f(x)=\frac1{1+25x^2}\) using equally spaced points
I'm sure you'll agree that these approximations are pretty terrible, and they get worse as more points are added. The high error towards 1 and -1 is called Runge's phenomenon, and was discovered in 1901 by Carl David Tolmé Runge.
All hope of finding a good polynomial approximation is not lost, however: by choosing the points more carefully, it's possible to avoid Runge's phenomenon. Chebyshev points (named after Pafnuty Chebyshev) are defined by taking the \(x\) co-ordinate of equally spaced points on a circle.
![](javascript:showlimage("chebdef.png"))
Eight Chebyshev points
The following GIF shows interpolating polynomials of the same function as before using Chebyshev points.
![](javascript:showlimage("runge-chebyshev.gif"))
Nice, we've found a polynomial that closely approximates the function... But I guess you're now wondering how well the Chebyshev interpolation will approximate other functions. To find out, let's try it out on the votes over time of my first round Big Internet Math-Off match.
![](javascript:showlimage("data.png"))
Scroggs vs Parker, 6-8 July 2018
The graphs below show the results of the match over time interpolated using 16 uniform points (left) and 16 Chebyshev points (right). You can see that the uniform interpolation is all over the place, but the Chebyshev interpolation is very close the the actual results.
![](javascript:showlimage("uniform.png"))
![](javascript:showlimage("chebyshev.png"))
Scroggs vs Parker, 6-8 July 2018, approximated using uniform points (left) and Chebyshev points (right)
But maybe you still want to see how good Chebyshev interpolation is for a function of your choice... To help you find out, I've wrote @RungeBot, a Twitter bot that can compare interpolations with equispaced and Chebyshev points.
Since first publishing this post, Twitter's API changes broke @RungeBot, but it lives on on Mathstodon: @RungeBot@mathstodon.xyz.
Just tweet it a function, and it'll show you how bad Runge's phenomenon is for that function, and how much better Chebysheb points are.
For example, if you were to toot "@RungeBot@mathstodon.xyz f(x)=abs(x)", then RungeBot would reply: "Here's your function interpolated using 17 equally spaced points (blue) and 17 Chebyshev points (red). For your function, Runge's phenomenon is terrible."
![](javascript:showlimage("runge-abs.jpg"))
A list of constants and functions that RungeBot understands can be found here.
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Hi Matthew, I really like your post. Is there a benefit of using chebyshev spaced polynomial interpolation rather than OLS polynomial regression when it comes to real world data? It is clear to me, that if you have a symmetric function your approach is superior in capturing the center data point. But in my understanding in your vote-example a regression minimizing the residuals would be preferrable in minimizing the error. Or do I miss something?
Benedikt
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2018-07-07
So you've calculated how much you should expect the World Cup sticker book to cost
and recorded how much it actually cost. You might be wondering what else you can do with your sticker book.
If so, look no further: this post contains 5 mathematical things involvolving your sticker book and stickers.
Test the birthday paradox
![](javascript:showlimage("birthdaysbrazil.jpg"))
Stickers 354 and 369: Alisson and Roberto Firmino
In a group of 23 people, there is a more than 50% chance that two of them will share a birthday. This is often called the birthday paradox, as the number 23 is surprisingly small.
Back in 2014 when Alex Bellos suggested testing the birthday paradox on World Cup squads, as there are 23 players in a World Cup squad. I recently discovered that even further back in 2012, James Grime made a video about the birthday paradox in football games, using the players on both teams plus the referee to make 23 people.
In this year's sticker book, each player's date of birth is given above their name, so you can use your sticker book to test it out yourself.
Kaliningrad
![](javascript:showlimage("kaliningrad.jpg"))
Sticker 022: Kaliningrad
One of the cities in which games are taking place in this World Cup is Kaliningrad. Before 1945, Kaliningrad was called Königsberg. In Königsburg, there were seven bridges connecting four islands. The arrangement of these bridges is shown below.
![](javascript:showlimage("7bridges.png"))
The people of Königsburg would try to walk around the city in a route that crossed each bridge exactly one. If you've not tried this puzzle before, try to find such a route now before reading on...
In 1736, mathematician Leonhard Euler proved that it is in fact impossible to find such a route. He realised that for such a route to exist, you need to be able to pair up the bridges on each island so that you can enter the island on one of each pair and leave on the other. The islands in Königsburg all have an odd number of bridges, so there cannot be a route crossing each bridge only once.
In Kaliningrad, however, there are eight bridges: two of the original bridges were destroyed during World War II, and three more have been built. Because of this, it's now possible to walk around the city crossing each bridge exactly once.
![](javascript:showlimage("kaliningrad-path.png"))
A route around Kaliningrad crossing each bridge exactly once.
I wrote more about this puzzle, and using similar ideas to find the shortest possible route to complete a level of Pac-Man, in this blog post.
Sorting algorithms
If you didn't convince many of your friends to join you in collecting stickers, you'll have lots of swaps. You can use these to practice performing your favourite sorting algorithms.
Bubble sort
In the bubble sort, you work from left to right comparing pairs of stickers. If the stickers are in the wrong order, you swap them. After a few passes along the line of stickers, they will be in order.
![](javascript:showlimage("bubble-sort-stickers.gif"))
Bubble sort
Insertion sort
In the insertion sort, you take the next sticker in the line and insert it into its correct position in the list.
![](javascript:showlimage("insertion-sort-stickers.gif"))
Insertion sort
Quick sort
In the quick sort, you pick the middle sticker of the group and put the other stickers on the correct side of it. You then repeat the process with the smaller groups of stickers you have just formed.
![](javascript:showlimage("quick-sort-stickers.gif"))
Quick sort
The football
![](javascript:showlimage("correct-football.jpg"))
Sticker 007: The official ball
Sticker 007 shows the official tournament ball. If you look closely (click to enlarge), you can see that the ball is made of a mixture of pentagons and hexagons. The ball is not made of only hexagons, as road signs in the UK show.
Stand up mathematician Matt Parker started a petition to get the symbol on the signs changed, but the idea was rejected.
If you have a swap of sticker 007, why not stick it to a letter to your MP about the incorrect signs as an example of what an actual football looks like.
Psychic pets
Speaking of Matt Parker, during this World Cup, he's looking for psychic pets that are able to predict World Cup results. Why not use your swaps to label two pieces of food that your pet can choose between to predict the results of the remaining matches?
![](javascript:showlimage("timber-football1.jpg"))
![](javascript:showlimage("timber-football2.jpg"))
Timber using my swaps to wrongly predict the first match
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@Matthew: Thank you for the calculations. Good job I ordered the stickers I wanted #IRN. 2453 stickers - that's more than the number you bought (1781) to collect all stickers!
Milad
@Matthew: Here is how I calculated it:
You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
@Matthew: Yes, I would like to know how you work it out please. I believe I have left my email address in my comment. It seems like a lot of stickers if you are just interested in one team.
Milad
@Milad: Following a similiar method to this blog post, I reckon you'd expect to buy 2453 stickers (491 packs) to get a fixed set of 20 stickers. Drop me an email if you want me to explain how I worked this out.
Matthew
Thanks for the maths, but I have one probability question. How many packs would I have to buy, on average, to obtain a fixed set of 20 stickers?
Milad
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2018-07-06
This is a post I wrote for round 1 of The Aperiodical's Big Internet Math-Off 2018, where Mathsteroids lost to MENACE.
A map projection is a way of representing the surface of a sphere, such as the Earth, on a flat surface. There is no way to represent all the features of a sphere on a flat surface, so if you want a map that shows a certain feature of the world, then you map will have to lose some other feature of the world in return.
To show you what different map projections do to a sphere, I have created a version of the game Asteroids on a sphere. I call it Mathsteroids. You can play it here, or follow the links below to play on specific projections.
Mercator projection
The most commonly used map projection is the Mercator projection, invented by Gerardus Mercator in 1569. The Mercator projection preserves angles: if two straight lines meet at an angle \(\theta\) on a sphere, then they will also meet at an angle \(\theta\) on the map. Keeping the angles the same, however, will cause the straight lines to no longer appear straight on the map, and the size of the same object in two different place to be very different.
The angle preserving property means that lines on a constant bearing (eg 030° from North) will appear straight on the map. These constant bearing lines are not actually straight lines on the sphere, but when your ship is already being buffeted about by the wind, the waves, and the whims of drunken sailors, a reasonably straight line is the best you can expect.
The picture below shows what three ships travelling in straight lines on the sphere look like on a Mercator projection.
![](javascript:showlimage("sphere--Mercator.png"))
To fully experience the Mercator projection, you can play Mathsteroids in Mercator projection mode here. Try flying North to see your spaceship become huge and distorted.
Gall–Peters projection
The Gall–Peters projection (named after James Gall and Arno Peters) is an area-preserving projection: objects on the map will have the same area as objects on the sphere, although the shape of the object on the projection can be very different to its shape on the sphere.
The picture below shows what three spaceships travelling in straight lines on a sphere look like on the Gall–Peters projection.
![](javascript:showlimage("sphere--Gall.png"))
You can play Mathsteroids in Gall–Peters projection mode here. I find this one much harder to play than the Mercator projection, as the direction you're travelling changes in a strange way as you move.
Azimuthal projection
![](javascript:showlimage("UNemblem.png"))
The emblem of the UN
An azimuthal projection makes a map on which the directions from the centre point to other points on the map are the same as the directions on the sphere. A map like this could be useful if, for example, you're a radio operator and need to quickly see which direction you should point your aerial to communicate with other points on the map.
The azimuthal projection I've used in Mathsteroids also preserves distances: the distance from the centre to the another points on the map is proportional to the actual distance on the sphere. This projection is used as the emblem of the UN.
The picture below shows three straight lines on this projection. You can play Mathsteroids in azimuthal mode here.
![](javascript:showlimage("sphere--azim.png"))
Craig retroazimuthal projection
A retroazimuthal projection makes a map on which the directions to the centre point from other points on the map are the same as the directions on the sphere. If you're thinking that this is the same as the azimuthal projection, then you're too used to doing geometry on flat surfaces: on a sphere, the sum of the angles in a triangle depends on the size of the triangle, so the directions from A to B and from B to A aren't as closely related as you would expect.
The Craig retroazimuthal projection was invented by James Ireland Craig in 1909. He used Mecca as his centre point to make a map that shows muslims across the world which direction they should face to pray.
The picture below shows what three spaceships travelling in a straight lines on a sphere looks like on this projection.
![](javascript:showlimage("sphere--Craig.png"))
You can play Mathsteroids in Craig retroazimuthal mode here to explore the projection yourself. This is perhaps the hardest of all to play, as (a) two different parts of the sphere overlap on the map, and (b) the map is actually infinitely tall, so quite a bit of it is off the edge of the visible game area.
Stereographic projection
The final projection I'd like to show you is the stereographic projection.
Imagine that a sphere is sitting on a 2D plane. Take a point on the sphere. Imagine a straight line through this point and the point at the top of the sphere. The point where this line meets the 2D plane is stereographic projection of the point on the sphere.
![](javascript:showlimage("stereographic3D.png"))
The stereographic projection
This projection (backwards) can be used to represent the every complex number as a point on a sphere: this is called the Riemann sphere.
To make Mathseteroids playable after this projection, I split the sphere into 2 hemisphere and projected each seperately to give two circles. You can play Mathsteroids in stereographic projection mode here. Three spaceships travelling in straight lines on this projection are shown below.
![](javascript:showlimage("sphere--stereographic.png"))
... and if you still don't like map projections, you can still enjoy playing Mathsteroids on an old fashioned torus. Or on a Klein bottle or the real projective plane. Don't forget to take a short break from playing to head over to The Aperiodical and vote (voting now closed).
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