# Blog

**2019-09-01**

This week, I've been in Cambridge for Talking Maths in Public (TMiP). TMiP is a conference for anyone involved in—or interested in getting involved
in—any sort of maths outreach, enrichment, or public engagement activity. It was really good, and I highly recommend coming to TMiP 2021.

The Saturday morning at TMiP was filled with a choice of activities, including a treasure punt (a treasure hunt on a punt) written by me. This post contains the puzzle from the treasure punt for
anyone who was there and would like to revisit it, or anyone who wasn't there and would like to give it a try. In case you're not current in Cambridge on a punt, the clues that you were meant to
spot during the punt are given behing spoiler tags (hover/click to reveal).

### Instructions

Each boat was given a copy of the instructions, and a box that was locked using a combination lock.

If you want to make your own treasure punt or similar activity, you can find the LaTeX code used to create the instructions and the Python code I used to check that the puzzle
has a unique solution on GitHub. It's licensed with a CC BY 4.0
licence, so you can resuse an edit it in any way you like, as long as you attribute the bits I made that you keep.

### The puzzle

Four mathematicians—Ben, Katie,
Kevin, and Sam—each have one of the four clues needed to unlock a great treasure.
On a sunny/cloudy/rainy/snowy (delete as appropriate) day, they meet up in Cambridge to go punting, share their clues, work out the code for the lock,
and share out the treasure. One or more of the mathematicians, however, has decided to lie about their clue so they can steal all the treasure for themselves.
At least one mathematician is telling the truth.
(If the mathematicians say multiple sentences about their clue, then they are either all true or all false.)

They meet at Cambridge Chauffeur Punts, and head North under Silver Street Bridge.
Ben points out a plaque on the bridge with two years written on it:

"My clue," he says, "tells me that the sum of the digits of the code is equal to the sum of the digits of the earlier year on that plaque (the year is 1702). My clue also tells me that at least one of the digits of the code is 7."

The mathematicians next punt under the Mathematical Bridge, gasping in awe at its tangential trusses, then punt along the river under King's College Bridge and past King's College.
Katie points to a sign on the King's College lawn near the river:

"See that sign whose initials are PNM?" says Katie. "My clue states that first digit of the code is equal to the number of vowels on that sign (The sign says "Private: No Mooring").
My clue also tells me that at least one of the digits of the code is 1."

They then reach Clare Bridge. Kevin points out the spheres on Clare Bridge:

"My clue," he says, "states that the total number of spheres on both sides of this bridge is a factor of the code (there are 14 spheres). My clue also tells me that at least one of the digits of the code is 2."
(Kevin has not noticed that one of the spheres had a wedge missing, so counts that as a whole sphere.)

They continue past Clare College. Just before they reach Garret Hostel Bridge, Sam points out the Jerwood Library and a sign showing the year it was built (it was built in 1998):

"My clue," she says, "says that the largest prime factor of that year appears in the code (in the same way that you might say the number 18 appears in 1018 or 2189).
My clue also says that the smallest prime factor of that year appears in the code. My clue also told me that at least one of the digits of the code is 0."

They then punt under Garret Hostel Bridge, turn around between it and Trinity College Bridge, and head back towards Cambridge Chauffeur Punts.

*Zut alors*, the lies confuse them and they can't unlock the treasure. Can you work out who is lying and claim the treasure for yourself?### The solution

The solution to the treasure punt is given below. Once you're ready to see it, click "Show solution".

### Similar posts

Christmas card 2018 | Christmas card 2017 | Christmas card 2016 | The Mathematical Games of Martin Gardner |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**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.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.

Drawing two curves in each square gives the 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.

The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.

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:

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:

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 rectangles | Polynomial | Solution |

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:

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

Dragon curves II | TMiP 2019 treasure punt | Christmas card 2018 | World Cup stickers 2018, pt. 3 |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2019-05-02**

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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.

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 |

### Similar posts

Christmas card 2017 | Christmas card 2016 | TMiP 2019 treasure punt | Christmas (2018) is over |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2018-12-17**

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**2017-12-18**

Just like last year, TD and I spent some time in November this year designing a puzzle Christmas card for Chalkdust.

The card looks boring at first glance, but contains 10 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s black and 2s orange will reveal a Christmassy picture.

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 converted to base 3 and coloured...

# | Answer (base 10) | Answer (base 3) | ||||||||

1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

4 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

5 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

6 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

7 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||

10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

- In a book with 116 pages, what do the page numbers of the middle two pages add up to?
- What is the largest number that cannot be written in the form \(14n+29m\), where \(n\) and \(m\) are non-negative integers?
- How many factors does the number \(2^6\times3^{12}\times5^2\) have?
- How many squares (of any size) are there in a \(15\times14\) grid of squares?
- You take a number and make a second number by removing the units digit. The sum of these two numbers is 1103. What was your first number?
- What is the only three-digit number that is equal to a square number multiplied by the reverse of the same square number? (The reverse cannot start with 0.)
- What is the largest three-digit number that is equal to a number multiplied by the reverse of the same number? (The reverse cannot start with 0.)
- What is the mean of the answers to questions 6, 7 and 8?
- How many numbers are there between 0 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6?
- What is the lowest common multiple of 52 and 1066?

### Similar posts

Christmas card 2016 | Christmas card 2018 | TMiP 2019 treasure punt | Christmas (2018) is over |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2019-01-04**

**10,000**and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6.

**2018-01-01**

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**2017-11-14**

A few weeks ago, I took the copy of MENACE that I built to Manchester Science Festival, where it played around 300 games against the public while learning to play Noughts and Crosses. The group of us operating MENACE for the weekend included Matt Parker, who made two videos about it. Special thanks go to Matt, plus
Katie Steckles,
Alison Clarke,
Andrew Taylor,
Ashley Frankland,
David Williams,
Paul Taylor,
Sam Headleand,
Trent Burton, and
Zoe Griffiths for helping to operate MENACE for the weekend.

As my original post about MENACE explains in more detail, MENACE is a machine built from 304 matchboxes that learns to play Noughts and Crosses. Each box displays a possible position that the machine can face and contains coloured beads that correspond to the moves it could make. At the end of each game, beads are added or removed depending on the outcome to teach MENACE to play better.

### Saturday

On Saturday, MENACE was set up with 8 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. I had only included one copy of moves that are the same due to symmetry.

The plot below shows the number of beads in MENACE's first box as the day progressed.

Originally, we were planning to let MENACE learn over the course of both days, but it learned more quickly than we had expected on Saturday, so we reset is on Sunday, but set it up slightly differently. On Sunday, MENACE was set up with 4 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. This time, we left all the beads in the boxes and didn't remove any due to symmetry.

The plot below shows the number of beads in MENACE's first box as the day progressed.

You can download the full set of data that we collected over the weekend here. This includes the first two moves and outcomes of all the games over the two days, plus the number of beads in each box at the end of each day. If you do something interesting (or non-interesting) with the data, let me know!

### Similar posts

MENACE | Building MENACEs for other games | MENACE in fiction | The Mathematical Games of Martin Gardner |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2017-11-22**

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