On the 10th of October we engaged in an activity in which thousands of students around the world were concurrently engaged.
First, pick a rectangle in the room (e.g. table, computer screen, piece of paper). Take a piece of string and make it the length of the perimeter of the rectangle plus one metre.
Now try and hold the string around the shape so that you have made a path of equal width around the shape. How wide is your path?
We did this practically and then resorted to pen and paper to find exact widths for different sized starting rectangles.
Amazingly we all found the same width …
We then tried the same puzzle but keeping the path the same width all the way round the shape:
Finally we tried with a circle
which led to us being able to understand the answer to this classic puzzle:
Imagine that the Earth is a perfect sphere and imagine we have a large piece of string tied around the equator.
We then add one meter to the length of the string so that now the string is hovering above the equator, still in a circle.
What is the size of the gap between the equator and the piece of string? Enough to fit a piece of paper under? A cat? A car?
You are given two lives to guess my number (which is between 1 and 100 inclusive).
If you guess too high you lose a life.
For example …
I’m secretly thinking of the number 42. If you guess 50 you lose a life. If you then guess 45 you lose your final life and you have failed.
After guessing 50 and losing a life you would then have to guess 1, guess 2, guess 3 and so on until you correctly guess 42. So this would work but would take a long time.
What strategy should you use to minimise the number of guesses it takes to guess my number? And what is the worst case scenario for how many guesses it could take?
If you are familiar with the two eggs problem it is essentially the same puzzle!
Have you met Happy Numbers? Start with any number you like and form a sequence by writing down the sum of the squares of the digits of each number to get the next number in the sequence. For example, from 31 you go to 10 because 3 squared plus 1 squared makes 10. Here is a ‘happy’ sequence:
31 -> 10 -> 1 -> 1 -> 1 …
Numbers are called ‘happy’ when their sequences, sooner or later, give repeated 1’s. We say 1 is a fixed point.
The number 25 is sad because, however long we go on with the sequence, it will never come to 1, it will just keep repeating the terms (89, 145, 42, 20, 4, 16, 37, 58) over and over again. We call this an 8-cycle or loop. Can you find some more happy numbers?
We were delighted to welcome ex Lycee student Al Sergeevo back to Maths Club. He did a fantastic interactive presentation on graph theory culminating in using Euler’s formula to derive the number of faces on a dodecahedron.
We played a two player strategy game today, using all the information from +plus magazine.
The game is played as follows. At the start of a game each player decides on their three colour sequence for the whole game. The cards are then turned over one at a time and placed in a line, until one of the chosen triples appears. The winning player takes the upturned cards, having won that “trick”. The game continues with the rest of the unused cards, with players collecting tricks as their triples come up, until all the cards in the pack have been used. The winner of the game is the player that has won the most tricks. An average game will consist of around 7 “tricks”.
e.g. Player 1 picks RRB, Player 2 picks RBB and see who wins.
The question we thought about was:
Is it just a game of chance, or if you are choosing second, could you improve your chance of winning?
There is a related game called Penny’s game, using Heads and Tails of a coin toss instead of playing cards. In this version, you just play until one person has won a “trick”. We worked out some of the odds given in the first table in this article.
We ended up having to sum a geometric series! Great fun!
This game is thanks to David Bedford at the BCME Conference in Warwick.
One person writes down a polynomial with positive integer coefficients. Call it f(x) They then choose an integer that it bigger than all their coefficients. Call it n. They then calculate f(n) and give just n and f(n) to the second person.
The second person should be able to name that polynomial! How?
Example:
Given only n=8 and f(n) = 6855 how could you work out that the polynomial was
?
It might help to think about an example when n=10 first.
Here is a cheatsheet for when you have a strategy. Can you write some Python code to work out the polynomial for you?