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2025-08-30
This week, I've been at Talking Maths in Public (TMiP) at the University of Warwick in near Coventry. 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 2027... But
as I'm one of the organisers, I'm a little biased.
The Saturday morning at TMiP was filled with a choice of activities, including a puzzle hunt written by me.
At the start/end point of the puzzle hung, there was a locked box with a combination lock. In order to work out the combination for the lock, you needed to find some clues hidden around
Coventry and solve a few puzzles.
Every team taking part was given a copy of these instructions.
Some people attended TMiP virtually, so I also made a version of the puzzle hunt that included links to Google Street View and photos from which the necessary information could be obtained.
You can have a go at this at mscroggs.co.uk/coventry-trail/remote. For anyone who wants to try the puzzles without searching through virtual Coventry,
the numbers that you needed to find are:
- Clue #1: \(a\) is 1931.
- Clue #2: \(b\) is 1956.
- Clue #3: \(c\) is 1434.
- Clue #4: \(d\) is 1949.
- Clue #5: \(e\) is 1620.
The solutions to the puzzles and the final puzzle are below. If you want to try the puzzles for yourself, do that now before reading on.
Puzzle for clue #1
154 is equal to 50625. The hundreds digit of 154 is 6.
The difference between the first and second digits of the code is the hundreds digit of \(15^a\) (ie 151931).
Puzzle for clue #2
If you write the numbers from 1 to 10000 in a huge triangle like this:
| 1 | |||||
| 2 | 3 | 4 | |||
| 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | ... |
... then 11 is written directly below 5. The second digit of the code is not the tens digit of the number written directly below \(b\) (ie directly below 1956).
Puzzle for clue #3
The area of largest quadrilateral that fits inside a circle with area 2π is 4.
The difference between the first and last digits of the code is the thousands digits of the area of the largest dodecagon that fits inside a circle with area \(c\)π (ie 1434π).
Puzzle for clue #4
There are 10 dominoes that can be made using the numbers 0 to 3 (inclusive):
| 0 | 0 |
| 0 | 1 |
| 0 | 2 |
| 0 | 3 |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 2 |
| 2 | 3 |
| 3 | 3 |
The sum of all the numbers on all these dominoes is 30.
The difference between the largest and smallest digits in the code is the units digit of the sum of all the numbers
on all the dominoes that can be made using the numbers 0 to \(d\) (ie from 0 to 1949) (inclusive).
Puzzle for clue #5
The number \(n\) has \(e\) digits (ie 1620 digits). All of its digits are 9. The last digit of the code is the hundreds digit of the sum of all the digits of \(n^2\).
The final puzzle
The final puzzle involves using the answers to the five puzzles to find the four digit code that
opens the box (and the physical locked box that was in the Transport Museum on
Saturday.
The five clues to the final code are:
- Clue #1: The difference between the first and second digits of the code is 3.
- Clue #2: The second digit of the code is not 4.
- Clue #3: The difference between the first and last digits of the code is 4.
- Clue #4: The difference between the largest and smallest digits in the code is 5.
- Clue #5: The last digit of the code is 5.
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2025-09-13
Great to do remotely, thanks! I think there is a typo in your solution for clue 4 although final answer is correct. Last sum prior to answer should read 1951 x 1/2 x 1949 x 1950 I think. Lizzie
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2024-02-20
Back in November, I wrote about making 2n-page zines.
Thanks to some conversations I had at Big MathsJam
in later November, I've been able to work out how many 128-page zines there are: 315434.
The insight
At Big MathsJam, Colin Beveridge pointed out
something he'd noticed about the possible zines: when drawing the line connecting the pages
in order, there were some line segments that were always included. For example, here are all of the possible
64-page zines:
![](javascript:showlimage("64zines.png"))
Every single one of these includes these line segments:
![](javascript:showlimage("64zines-always.png"))
Colin conjectured that for a zine of any size, a pattern like this of alternative horizontal segments
must always be included. He was close to justifying this, and since MathsJam I've been able to fill
in the full justificication.
The justificiation
First, consider the left-most column of pages. They must be connected like this:
![](javascript:showlimage("64zines-leftcol.png"))
If they were connected in any other way, there would be two vertical connections in a row,
which would create a page that is impossible to open (as every other connection must be a horizontal
that ends up in the spine). Additionally, the horizontal lines in this diagram must all be in the
spine (as otherwise we again get pages that cannot be opened).
Next, consider a horizontal line that's in the spine (shown in red below), and we can look
at all the possible ways to draw the line through the highlighted page, paying particular
attention to the dashed blue line:
![](javascript:showlimage("64zines-highlighted.png"))
The six possible ways in which the line could travel through the highlighted page are:
![](javascript:showlimage("64zines-options.png"))
The three options in the top row do not give a valid zine: the leftmost diagram has two vertical
connections in a row (leading to pages that do not open). The other two diagrams in the top row
have the horizontal line that we know is in the spine, followed by a horizontal line not in the
spine, then a vertial line: this vertical line should be in the spine, but as it is vertical
it cannot be (without making a page that doesn't open).
In each of the diagrams in the bottom row, the connection shown in dashed blue
is included and must be in the spine: in the leftmost diagram, the horizontal line that we know is in the spine
is followed by a horizontal not in the spine, then the horizinal in the dashed blue position
that must therefore be in the spine. The othe other two diagrams in the bottom row,
the dashed blue position is connected to a vertical line: this means that the dashed blue connection
must be in the spine (as otherwise the vertical would cause a page that doesn't open).
Overall, we've now shown that the leftmost column of lines must always be included and
must all be in the spine; and for each horizontal line in the spine, the line to the right of it
after a single gap must also be included and in the spine. From this, it follows that all the horizontal
lines in Colin's pattern must always be included.
Calculating the number of 128-page zines
Now that I knew that all these horizonal lines are always included, I was able to update
the code I was using to find all the possible zines
to use this. After a few hours, it had found all 315434 possibilites. I was very happy to get this
total, as it was the same as the number that
Luna (another attendee of Big MathsJam) had calculated but wasn't certain was correct.
The sequence of the number of 2n-page zines,
including the newly calculated number,
is now published on the OEIS.
I think calculating number of 256-page zines is still beyond my code though...
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2023-11-03
I showed off an unfinished version of the content in this blog post in the first
Finite Group live stream in October. I'll be showing off other upcoming things and other
content at future Finite Group events. If you'd like to watch this, you can sign up to our Patreon at
finitegroup.co.uk.
A few months ago at Cheltenham Science Festival, Hana Ayoob showed me how to make an 8-page zine.
If you've not make one before, I recommend following these steps now.
Take a piece of A4 paper, and fold it 3 times like this:
![](javascript:showlimage("8zine1.png"))
Unfold all the folds except the first one and cut along the red line:
![](javascript:showlimage("8zine2.png"))
After unfolding, your paper should look like this:
![](javascript:showlimage("8zine3.png"))
Fold the paper in half the other way:
![](javascript:showlimage("8zine4.png"))
Push the two ends of the paper together so that the front of the middle bit of paper comes forward,
and the back goes backward:
![](javascript:showlimage("8zine5.png"))
Finally, flatten all the pages and you have your zine:
![](javascript:showlimage("8zine6.png"))
You're now free to write or draw whatever you like on the 8 pages of your zine.
If you'd like to fold a zine that already has content printed on it, you can
print the instructions for the TMiP puzzle hunt that I wrote earlier this year, or you can print this blog post off in zine format.
A 16-page zine
If you're anything like me, you'll already be wondering if it's possible to do some more folding
and cutting to make a 16-page zine from a piece of A4. After some trial and error, I found that
you can if you fold the paper 4 times then cut along these red lines:
![](javascript:showlimage("16zine1.png"))
(This time, I've drawn the diagram with a portrait piece of paper, as this leads to a portrait zine. For an 8-page zine, a landscape piece of paper led to a portrait zine.)
Instead of drawing which lines we need to cut, we can represent our two zines so far by drawing
lines that connect the pages in order:
![](javascript:showlimage("8and16zines.png"))
During my experimentation, I saw that every second connection between pages must be horizontal.
These horizontal connections end up in the spine of the zine and allow the pages to turn.
If you can't visualise why the pages won't turn if this condition doesn't hold, try making a 16-page
zine like this:
![](javascript:showlimage("failed16zine.png"))
32-page zines
Now that we can make a 16-page zine, the obvious question is: can we make a 32-page zine?
To answer this, we need to look for lines that go through all
32 pages where every second connection is a horizontal. I wrote a Python script to look for these and it
found 3:
![](javascript:showlimage("32zines.png"))
The code also told me that the 8- and 16-page zines we know about are the only 8- and 16-page
zines. I'm only counting the zines where the pages in the final zine are portrait, and have the same side ratio a the original piece of A4 paper:
there will be other possible zines where the pages are landscape that we aren't counting.
2n-page zines
Now that we have 8-page, 16-page, and 32-page zines, we can look for patterns that we
can generalise to make a 2n-page zine. This generalisation was the first I
came up with:
![](javascript:showlimage("2nzines.png"))
One possible 2n-page zine for n=3 to 12 (click to enlarge)
There are, of course, many other generalisations that you could come up with.
How many zines?
While working towards our generalised zine, you may have started pondering another question:
for any given n, how many different 2n-page zines can be made?
For n=2, there's only one way to make a zine (fold the paper twice).
For n=3 to 6, we've already seen that there are 1, 1, and 3 ways to make zines. The code
I wrote was also able to tell me that there are 31 possible 64-page zines:
All 31 possible 64-page zines (click to enlarge)
So the start of the sequence of the number of possible zines is: 1, 1, 1, 3, 31.
The number of different possible paths to check increases very quickly as we increase
n, so I was unable to compute the next term in a reasonable amount of time.
I've submitted this sequence to the OEIS. Let's
hope someone is able to work out the next term. If you're that someone, let me know!
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This is hilarious. I was looking up ways to fold zines beyond the typical 8-pager and found this. Thanks for the laughs. I will test some of these out tomorrow.×1
Lou
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2023-09-02
This week, I've been at Talking Maths in Public (TMiP) in Newcastle. 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 2025.
The Saturday morning at TMiP was filled with a choice of activities, including a puzzle hunt written by me: the Tyne trial.
At the start/end point of the Tyne trial, there was a locked box with a combination lock. In order to work out the combination for the lock, you needed to find some clues hidden around
Newcastle and solve a few puzzles.
Every team taking part was given a copy of these instructions.
Some people attended TMiP virtually, so I also made a version of the Tyne trial that included links to Google Street View and photos from which the necessary information could be obtained.
You can have a go at this at mscroggs.co.uk/tyne-trial/remote. For anyone who wants to try the puzzles without searching through virtual Newcastle,
the numbers that you needed to find are:
- Clue #1: \(a\) is 9.
- Clue #2: \(b\) is 5.
- Clue #3: \(c\) is 1838.
- Clue #4: \(d\) is 1931.
- Clue #5: \(e\) is 1619.
- Clue #6: \(f\) is 48.
- Clue #7: \(g\) is 1000.
The solutions to the puzzles and the final puzzle are below. If you want to try the puzzles for yourself, do that now before reading on.
Puzzle for clue #2: Palindromes
We are going to start with a number then repeat the following process: if the number you have is a palindrome, stop;
otherwise add the number to itself backwards.
For example, if we start with 219, then we do: $$219\xrightarrow{+912}1131\xrightarrow{+1311}2442.$$
If you start with the number \(10b+9\) (ie 59), what palindrome do you get?
(If you start with 196, it is unknown whether you will ever get a palindrome.)
Puzzle for clue #3: Mostly ones
There are 12 three-digit numbers whose digits are 1, 2, 3, 4, or 5 with exactly two digits that are ones.
How many \(c\)-digit (ie 1838-digit) numbers are there whose digits are 1, 2, 3, 4, or 5 with exactly \(c-1\) digits (ie 1837) that are ones?
Puzzle for clue #4: is it an integer?
The largest value of \(n\) such that \((n!-2)/(n-2)\) is an integer is 4. What is the largest value of \(n\) such that
\((n!-d)/(n-d)\) (ie \((n!-1931)/(n-1931)\)) is an integer?
Puzzle for clue #5: How many steps?
We are going to start with a number then repeat the following process:
if we've reached 0, stop; otherwise subtract the smallest prime factor of the current number.
For example, if we start with 9, then we do: $$9\xrightarrow{-3}6\xrightarrow{-2}4\xrightarrow{-2}2\xrightarrow{-2}0.$$ It took 4 steps to get to 0.
What is the smallest starting number such that this process will take \(e\) (ie 1619) steps?
Puzzle for clue #6: Four-digit number
I thought of a four digit number. I removed a
digit to make a three digit number, then added my two numbers together.
The result is \(200f+127\) (ie 9727). What was my original number?
Puzzle for clue #7: Dice
If you roll two six-sided fair dice, the most likely total is 7. What is the most likely total if you rolled \(1470+g\) (ie 2470) dice?
The final puzzle
The final puzzle involves using the answers to the puzzles to find the four digit code that
opens the box (and the physical locked box that was in the library on
Saturday. To give hints to this code, each clue was given a "score".
The score of a number is the number of values of \(i\) such that the \(i\)th digit
of the code is a factor of the \(i\)th digit of the number.
For example, if the code was 1234, then the score of the number 3654 would be 3 (because
1 is a factor of 3; 2 is a factor of 6; and 4 is a factor of 4).
The seven clues to the final code are:
- Clue #1: 6561 scores 2 points.
- Clue #2: 1111 scores 0 points.
- Clue #3: 7352 scores 1 points.
- Clue #4: 3562 scores 1 points.
- Clue #5: 3238 scores 1 points.
- Clue #6: 8843 scores 1 points.
- Clue #7: 8645 scores 3 points.
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