Modelling Coronavirus on Geogebra

Ben Sparks gives instructions to do some disease modelling on Geogebra – so you can see the maths behind the government advice to “flatten the curve”. Geogebra is available here. When you try and recreate the applet, don’t worry if the first three numbers appear as sliders, it will still work fine.

Also, see the message below from the Think Maths website we have enjoyed in the past:

“Matt Parker has launched Matt Parker’s Maths Puzzles! Once a week we can now look forward to a puzzle video from Matt on his Stand-Up Maths YouTube Channel.

Each week Matt will give us a puzzle and pose a question. Viewers can submit their solution to that question online to receive points and appear in a puzzle participants league table

Matt will be awarding hilarious virtual prizes when participants reach particular point milestones.

The first puzzle video is here. Submit your answer here: www.think-maths.co.uk/table-puzzle

We aim for puzzle videos to be released on a Wednesday afternoon UK time, with the deadline for submissions the Tuesday of the following week at 11:59pm UK time

Sign up to recieve an alert email when puzzle videos come out here: https://www.think-maths.co.uk/puzzles-sign-up

On Fridays we will post a solution video to the previous week’s puzzle and the updated league table on this page: www.think-maths.co.uk/maths-puzzles

Take care everyone.

Age brainteasers

Here are a couple of age puzzles from David Pleacher’s great site. Answers are on there too.

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During a recent census, a man told the census taker that he had 3 children.
When asked their ages, he replied, “The product of their ages is 72.”
“The sum of their ages is the same as my house number.”
The census taker ran to the door and looked at the house number.
“I still can’t tell,” she complained.
“Oh, that’s right. I forgot to tell you that the oldest one likes apple pie.”
The census taker promptly wrote down the ages of the three children.
How old are they?

****

Edie and Dave were talking when they saw three people coming toward them.
“I wonder how old they are,” said Edie.
Dave replied, “I know them!
The product of their ages is 2,450 and the sum of their ages is twice your age.”
“That’s all well and good,” said Edie, “but I need more information.”
“Oh yes,” said Dave.
“Well, I am older than any of the three.”
“Now, I can figure their ages,” said Edie.

How old are the three?

****

In case they were too easy here is a fiendish one to try by John H. Conway.

Last night I sat behind two wizards on a bus, and overheard the following: A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.” B: “How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?” A: “No.” B: “Aha! AT LAST I know how old you are!” Now what was the number of the bus?

Here is a paper which discusses the puzzle and solution.

Lorenz

A half term trip to Bletchley Park inspired the following activity …

Task 1

Go to
https://billtuttememorial.org.uk/codebreaking/teleprinter-code/

Read about teleprinter code and the rules of addition.

Task 2

In the alphabet in the link above, I and N are incorrect.

Can you use their addition table to work out what I and N should be? Remember that same symbols added make a dot, and different symbols make a cross.

Here is an addition table in alphabetical order.

Task 3

Code HELLO with the key ANQPC.
How would you get back to HELLO?

Solution here

Task 4

Read about the Tiltman break here

Task 5

Try it out for yourself!!

Above are two messages sent with the same key. One has been abbreviated after the operator was asked to send it again. Your crib is that it starts MESSAGE NUMBER (of course with a 9 in the middle!). You also know it is a weather report.  

If you add together the two messages letter by letter (using the table) you will end up with the two messages added together, because for:

Message1 +key + Message2 + key

the keys will cancel out and it will be Message1 + Message2

So if you can guess it starts MESSAGE9NUMBER then you can add this to the sum of the two messages and as they start to be different you can work out each one …

Four fours

Try to make all the numbers from 1 to 10 using exactly four fours and any number of the operators + – x / 

e.g. you can make 7 by doing 4 + 4 + 4/4

Can you make the rest?

If you are also allowed to use square root, factorial (!) and double factorial (!!) you can make the numbers 11 – 100.

e.g. 4! is 4x3x2x1 so you could make 28 by doing 4! + 4 + 4 – 4

Double factorial multiplies even or odd numbers like this:

4!! = 4 x 2

8!! = 8 x 6 x 4 x2

7!!= 7 x 5 x 3 x 1

Here is a full list of solutions if you get stuck, and watch the amazing video by Alex Bellos showing a solution for any number if you are allowed to use logarithms.

Lychrel numbers

Palindromes are words or numbers that read the same back to front. e.g. ABBA, racecar, 676, 128821, and a recent date 02/02/2020.

Puzzle

Take a positive integer. Reverse the digits to get a new number. Add the two numbers together.

Repeat this process until you get a palindrome.

Some numbers end up in a palindrome quite quickly e.g. 57. 57+75=132, 132+231=363.

Some numbers take a long time – 89 takes an unusually large 24 steps (the most of any number under 10,000 that is known to resolve into a palindrome according to Wikipedia) to reach the palindrome 8,813,200,023,188.

Can you find a starting number that doesn’t end up at a palindrome? Nobody know if this is possible.

You could use this applet that we wrote in maths club to look for one

It is interesting to look for Lychrel numbers in other bases – what happens in binary?

Gale-Shapley

We were very lucky to have ex Lycee student and maths club attendee Sophia Sergeeva come today to run a session on the Gale-Shapley Stable Marriage Problem. It was absolutely fascinating and delivered brilliantly.

The Problem

This problem looks at matchings: one to one correspondences between sets.

Suppose that we are looking at a strictly heterosexual space (e.g. Tinder) with four boys and four girls. A matching between the boys and girls is sought for. 

Let each of the girls have a list of preferences for the boys (from best to worst) and vice versa.

Let’s name the boys MA MB MC MD and the girls FA FB FC FD and list their preferences as follows.

The challenge is to match them up to maximise their happiness such that the system is “stable”. Stable in this sense means that no two people will want to run away together. For example if there exists two couples F1M1 and F2M2 then if F1 prefers M2  to M1 and M2 prefers F1 to F2  then this is unstable as they will want to run away. Note that it requires both to be true, if there is a couple for which one is top of the list for the other, but the other is bottom then it doesn’t matter as they can’t run away! 

Challenge 1

Find a stable matching for the boys and girls above

The Algorithm

Does a stable matching always exist? Yes, and there is an algorithm to find it. 

In each round of the this first very “traditional” algorithm, each unengaged man proposes to his next favourite woman who hasn’t rejected him. Each woman, if she has one or more suitors, becomes engaged to her favourite suitor and rejects all the others. Rounds continue until men have no one to propose to or are engaged. In subsequent rounds, females can break their engagement if someone they prefer proposes.

In the first stage of this algorithm for our list of preferences, the first round would work like this: 

MA  – FB

MB  – FC

MC  – FB

MD  – FD

FC and FD have only one suitor so would accept them. FB has a choice so would choose MC as he is further up her list of preferences. 

Challenge 2

Carry out the algorithm on the original list to find a stable matching.

Challenge 3

Which group are on average happier in this stable matching, the boys or the girls? Does it make a difference if the females propose? 

Challenge 4

Is there a way that the girls could “lie” in the algorithm to maximise their happiness even if the boys were proposing?!

Challenge 5

Now think of the general case. Does the algorithm always stop? Does it always produce a stable matching? Can you prove it?

Challenge 6

There doesn’t always exist a stable matching for non-hetrosexual preferences. Can you think of a list of preferences of 4 people such that no stable marriage exists? What about 3?

Applications

The Gale-Shapley algorithm is used in real life to solve problems like assigning university places and assigning doctors to hospitals.  Thanks again to Sophia for introducing us to this fascinating problem and algorithm.

4,3,2

Today we played another game from mathemagician Andrew Jeffrey’s great book. Andrew very kindly donated the proceeds of this book to SAMI. 

This game using the cards from Ace (one) up to 9 from a deck of cards.

  • The cards are shuffled and each player is dealt 9 cards.
  • The players then have to arrange their cards to make a 4-digit number, a 3-digit number and a 2-digit number and put the cards face down on the table.
  • Then the players show their numbers and compare them to each other.
  • The player with the biggest 2-digit number receives 2 points, the one with the biggest 3-digit number receives 3 points and the one with the biggest 4-digit number receives 4 points. 
  • If there’s a tie then the points are split between the players.
  • Play five times and see who is the winner …

Can you find a good strategy for this game?

Red black card game

Consider a single player card game with a standard 52 card deck.

You shuffle the cards and turn over the top card.

If you turn over a black card, you win £1. If you turn over a red card, you lose £1.

You can keep turning over cards as long as you want to.

It is impossible to lose money in this game; if you have turned over more red cards than black cards, you can just keep turning over cards until the end of the deck to end up at net £0. If you stop turning over cards before reaching the end of the deck, you keep the money that you have won up until that point.

Is there an optimal strategy for this game?

We played several games of this in maths club just with 10 cards (5 red and 5 black), keeping track of our winnings to start to guess at the best strategy.

Kaprekar’s number

Mr D. R. Kaprekar was an Indian school maths teacher who loved playing with numbers. See if you can follow these steps to find out what he discovered …

Choose a four digit number where the digits are not all the same (that is not 1111, 2222,…).

Rearrange the digits to get the largest and smallest numbers these digits can make.

Finally, subtract the smallest number from the largest to get a new number, and carry on repeating the operation for each new number.

What do you notice?

Does something similar happen for 3 digit numbers? Can you prove it?

Here are pdfs of the problem in English and French.

Try it out here