Factors and Multiples Chain

Today we worked on this lovely puzzle from Nrich.

Choose a starting number from a 1-100 square and cross it out.

Then choose a factor or multiple of that number.

Keep crossing out factors or multiples of the last number in the chain.

For example, Charlie started with 60, 30, 6, 96, 16, 32, 8, 56, 7, 21, 42,…

What’s the longest chain you can make?

There is an interactive place to play here

When you are using the activity make sure you only have a bracket at the start and the end … this attempt isn’t quite right:

There are two chains, rather than one continuous one.

But it can be fixed by swapping the 60,90,45 and 15 around:

Now it is a chain of 38 steps

Email Mrs Fleming on [email protected] if you can do better than 38 steps …

Sprouts

In honour of John H. Conway, today we present a game that he co-invented with Michael S. Paterson while they were both at Cambridge.

It is called Sprouts, and the rules are summarised by Nrich here.

Anyone can play, so find someone in your house and play a few games, then try and discover some of the maths behind it by working through this article.

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.

****

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.