Blog
TMiP 2025 puzzle hunt
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.
×1 (Click on one of these icons to react to this blog post)
You might also enjoy...
Comments
Comments in green were written by me. Comments in blue were not written by me.
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
Add a Comment