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TMiP 2021 puzzle hunt
2021-09-25
A few weeks ago, I (virtually) went to 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 2023.
The Saturday morning at TMiP was filled with a choice of activities, including a puzzle hunt written by me. Each puzzle required the solver to first find a clue hidden in
the conference's Gather-Town-powered virtual Edinburgh (built by the always excellent Katie Steckles), then solve the puzzle to reveal a clue to the final code. Once the final code was found, the solvers could enter
a secret area in the Gather Town space.
The puzzles for the puzzle hunt can be found at mscroggs.co.uk/tmip. For anyone who doesn't have access to the Gather Town space, the numbers that are hidden in the space are:
- Puzzle 1: The mathematician (Thomas Bayes) died in 1761.
- Puzzle 2: The sign claims that Arthur's seat is 1288 links tall.
- Puzzle 3: The sign claims that the maximum capicity of the museum is 2449 people.
- Puzzle 4: Dynamic Earth became a charitable trust in 94.
- Puzzle 5: The 102 bus goes to Dumfries.
The solutions to the five puzzles, and the final puzzle are below. If you want to try the puzzles for yourself, do that now before reading on.
Puzzle 1: The strange shop
A shop has a very strange pricing model. If you buy \(k\) items, then the price (in pence) is decided as follows:
- If \(k\) is prime, then the price is \(k\) pence.
- If \(k\) is not prime, then double \(k\) and add one:
- If the result is prime, that is the price.
- If the result is not prime, keep doubling and adding one until a prime is reached.
You enter the shop with 1761 pence and buy 28 items.
How many pence do you leave the shop with?
Fun fact: If you try to buy 509202 items from the shop, then the shopkeeper cannot work out a price,
as a prime is never reached. It is currently unknown if this is the smallest number of items that
this is true for.
Puzzle 2: The homemade notebook
You make a homemade notebook with 1288 pages:
You take a stack of 1288/4 pieces of paper and fold the entire stack in half so that
each piece of paper makes four pages in the notebook. You number the pages:
you write the number 1 on the front cover, 2 on the inside front cover, and so on
until you write 1288 on the back cover.
While you are looking for your stapler, a strong wind blows the pieces of paper all
over the floor. You pick up one of the pieces of paper and add up the two numbers
you wrote on one side of it.
What is the largest total you could have obtained?
Puzzle 3: The overlapping triangles
You draw three circles that all meet at a point:
You then draw two triangles. The smaller red triangle's vertices are the centres of the circles. The larger blue triangle's vertices are at the points on each circle diametrically opposite the point where all three circles meet:
The area of the smaller red triangle is 2449.
What is the area of the larger blue triangle?
The odd factors
You write down the integers from 94+1 to 2×94 (including 94+1 and 2×94). Under each number, you write down its largest odd factor*.
What is the sum of all the odd factors you have written?
* In this puzzle, factors include 1 and the number itself.
Hint: Doing what the puzzle says may take a long time. Try doing this will some smaller values than 94 first and see if you can spot a shortcut.
The sandwiched quadratic
You know that \(f\) is a quadratic, and so can be written as \(f(x)=ax^2+bx+c\) for some real numbers \(a\), \(b\), and \(c\);
but you've forgetten exactly which quadratic it is. You remember that for all real values of \(x\), \(f\) satisfies
$$\tfrac{1}{4}x^2+2x-8\leqslant f(x)\leqslant(x-2)^2.$$
You also remember that the minimum value of \(f\) is at \(x=0\).
What is f(102)?
The final puzzle
The final puzzle involves using the answers to the five puzzles to find a secret four digit passcode is made up of four non-zero digits. To turn them into clues,
the answers to each puzzle were scored as follows:
Each digit in an answer that is also in the passcode and in the same position in both scores two points; every digit in the answer that is also in the passcode but in a different position scores 1 point. For example, if the passcode was 3317, then:
- 4686 would score 0 points (no digits correct);
- 4676 would score 1 point (7 is a correct digit but in the wrong position);
- 4616 would score 2 points (1 is a correct digit in the correct position);
- 4343 would score 3 points (one of the 3s is a correct digit in the correct position, the other 3 is a correct digit in the wrong position);
- 7333 would score 4 points (one of the 3s is a correct digit in the correct position, the 7 and one of the other 3s are correct digits in wrong positions. Note that the third three scores nothing);
- and 1733 would score 4 points (all four digits are correct digits in wrong positions).
The five clues to the final code are:
- Puzzle 1: 1298 scores 3 points.
- Puzzle 2: 1289 scores 3 points.
- Puzzle 3: 9697 scores 3 points.
- Puzzle 4: 8836 scores 1 point.
- Puzzle 5: 5198 scores 4 points.
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Comments
Comments in green were written by me. Comments in blue were not written by me.
⭐ top comment (2021-09-27) ⭐
@Dan: Well spotted, I've edited the postMatthew
×2 ×4 ×3 ×3 ×3 Small nitpick on problem 1 fun fact. I think you meant 509202. 509203 is already prime so the price would be 509203. The way you set up the problem (2a_n+1) only gets to (k*2^n-1) if you start with k-1, so your k needs to be one smaller than the Wikipedia's k.×3 ×3 ×3 ×2 ×2
Dan
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