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 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:
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:
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.

Show solution

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?

Show solution

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?

Show solution

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.

Show solution

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)?

Show solution

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:
The five clues to the final code are:

Show solution

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Comments

Comments in green were written by me. Comments in blue were not written by me.
@Dan: Well spotted, I've edited the post
Matthew
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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.
Dan
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