# 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.

**2019-01-01**

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.

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.

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

### 12 December

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.

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).

### 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:
F

**red**Metcalfe, J**o Range**r, Bo**b Lue**y, Me**g Reen**y, and Ki**p Urple**s. 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.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |

Day |

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 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 Luey | Fred Metcalfe | Jo Ranger | Kip Urples | Meg 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 | 1:21–2:52Stealing 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:

1 | Sarah Brook |

2 | Mihai Zsisku |

3 | Bhavik Mehta |

4 | Peter Byrne |

5 | Martin Harris |

6 | Gert-Jan de Vries |

7 | Lyra |

8 | James O'Driscoll |

9 | Harry Poole |

10 | Albert 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.

### Similar posts

Christmas card 2018 | Christmas (2018) is coming! | Christmas (2017) is over | Christmas card 2017 |

### Comments

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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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**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—Jo Ranger, Bob Luey, Kip Urples, Meg Reeny, and Fred Metcalfe—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**!The winners will also receive one of these medals:

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!

### Similar posts

Christmas (2018) is over | Christmas card 2018 | Christmas (2017) is over | Christmas card 2017 |

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**2018-05-02**

Every morning just before 7am, one of the Today Programme's presenters reads out a puzzle. Yesterday, it was this puzzle:

In a given month, the probability of a certain daily paper either running a story about inappropriate behaviour at a party conference or running one about somebody's pet being retrieved from a domestic appliance is exactly half the probability of the same paper containing a photo of a Tory MP jogging. The probability of no such photo appearing is the same as that of there being a story about inappropriate behaviour at a party conference. The probability of the paper running a story about somebody's pet being retrieved from a domestic appliance is a quarter that of its containing a photo of a Tory MP jogging. What are the probabilities the paper will (a) run the conference story, (b) run the pet story, (c) contain the jogging photo?

I'm not the only one to notice that some of Radio 4's daily puzzles are not great.
I think this puzzle is a great example of a terrible puzzle. You can already see the first problem with it: it's long and words and very hard to follow on the radio.
But maybe this isn't so important, as you can
read it here after it's been read out.

Once you've done this, you can re-write the puzzle as follows:
there are three news stories (\(A\), \(B\) and \(C\)) that the newspaper might publish in a month. We are given the following information:

$$\mathbb{P}(A\text{ or }B)=\tfrac12\mathbb{P}(C)$$
$$1-\mathbb{P}(C)=\mathbb{P}(A)$$
$$\mathbb{P}(B)=\tfrac14\mathbb{P}(C)$$
To solve this puzzle, we need use the formula \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).
These Venn diagrams justify this formula:

Using the information we were given in the question, we get:

\begin{align}
\tfrac12\mathbb{P}(C)&=\mathbb{P}(A\text{ or }B)\\
&=1-\mathbb{P}(C)+\tfrac14\mathbb{P}(C)-\mathbb{P}(A\text{ and }B)\\
\mathbb{P}(C)&=\tfrac45(1-\mathbb{P}(A\text{ and }B)).
\end{align}
At this point we have reached the second problem with this puzzle: there's no answer unless we make some extra assumptions, and the question doesn't make it clear what we can assume.
But let's give the puzzle the benefit of the doubt and try some assumptions.

### Assumption 1: The events are mutually exclusive

If we assume that the events \(A\) and \(B\) are mutually exclusive—or, in other words, only one of these two articles can be published,
perhaps due to a lack of space—then we can use the fact that

$$\mathbb{P}(A\text{ and }B)=0.$$
This means that
\(\mathbb{P}(C)=\tfrac45\),
\(\mathbb{P}(A)=\tfrac15\), and
\(\mathbb{P}(B)=\tfrac15\). There's a problem with this answer, though: the three probabilities add up to more than 1.

This wouldn't be a problem, except we assumed that only one of the articles \(A\) and \(B\) could be published.
The probabilities adding up to more than 1 means that either \(A\) and \(C\) are not mutually exclusive or \(A\) and \(B\) are not mutually exclusive,
so \(C\) could be published alongside \(A\) or \(B\). There seems to be nothing special about the three news stories to mean that only some combinations
could be published together, so at this point I figured that this assumption was wrong and moved on.

Today, however, the answer was posted, and
this answer was given (without an working out). So we have a third problem with this puzzle: the answer that was given is wrong, or at the very best
is based on questionable assumptions.

### Assumption 2: The events are independent

If we assume that the events are independent—so one article being published doesn't affect whether or not another is published—then
we may use the fact that

$$\mathbb{P}(A\text{ and }B)=\mathbb{P}(A)\mathbb{P}(B).$$
If we let \(c=\mathbb{P}(C)\), then we get:

\begin{align}
\tfrac12c&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\\
&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)\\
&=1-c+\tfrac14c-\tfrac14(1-c)c\\
\tfrac14c^2-\tfrac32c+1&=0.
\end{align}
You can use your favourite formula to solve this to find that \(c=3-\sqrt5\), and therefore
\(\mathbb{P}(A)=\sqrt5-2\) and
\(\mathbb{P}(B)=\tfrac34-\tfrac{\sqrt5}4\).

In this case, our assumption appears to be more reasonable—as over the course of a month the stories published by a paper probably don't have
much of an effect on each other—but we have the fourth, and probably biggest problem with the puzzle: the question and answer are not interesting or surprising, and
the method is a bit tedious.

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**2018-05-03**

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