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TMiP 2023 puzzle hunt
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 five 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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