Lycee Francais SAMI Maths Club

Thanks to Barasa in Kenya for telling us about this one …

Challenge 1

Take 10 cards from a deck – A,2,3, … 10. Ace represents 1.

Arrange the cards in a pile face down so that when you show the first card it is the Ace. Remove this card. Then count to two putting the first card on the bottom of the pile. The second card should be the two. Show it and remove it. Then count to three, each time you say a number you should be putting a card on the bottom and then when you get to 3 you show the 3 and remove it from the pile.

Arrange the cards so that this works for every number from A to 10. By the time you get to 9 you should only have two cards left in your hand! Cycle through them counting 1, 2, 3, … 9 and show the 9. And then you just have 10 left and the trick is over.

Challenge 2

Try to arrange the cards to perform the same trick except you don’t remove the card when you show it. The card is just placed at the bottom of the pile and you start counting the next number.

You should find out that this is impossible for 10 cards. Is there a number of cards that it would work for?

We tried a dice puzzle today set a while back by Matt Parker.

You only have the tool above – three dice inside a cube.

But you want to play a game that requires throwing a single dice and then a game that requires throwing two dice and adding their score together. Is there a way that you can use the sum of these three dice to simulate both of these scenarios?

Task 1

First work out the distribution of numbers you can score with three dice

Can you map this in a nice way to the distribution for 1 dice?

Task 2

How about for 2 dice?

Might not be able to do it using (only) the sum …

Printable worksheet here.

Succinct solution here or watch the full brilliant video by Matt.

Today we worked on an absorbing puzzle that you can play online here:

https://www.cleverlearning.co.uk/blogs/blogConwayInteractive.php

The Irish logarithm is an algorithm invented by Percy Ludgate in 1909 for multiplying single-digit numbers. The idea was to program a computer to do these calculations.
The algorithm uses two tables to perform the multiplication. With just these tables you can calculate products up to 9×9 simply by adding two numbers together.

Here are blank versions of the two tables below (and as a pdf).

Our challenge was to fill in these tables. Here is the start of an attempt that would not work …

1 x 2 would be 1+2 so the correct answer of 2 needs to be in the box called 3.

1×3 would be 1+3 so the correct answer of 3 needs to be in the box called 4.

2×3 would be 2+3 so the correct answer of 6 needs to be in the box called 5.

So far so good ….

But … 2×2 would be 2+2 so the correct answer of 4 needs to go in the box called 4. But the box called 4 already has a 3 in it, so our initial choice of numbers is a bad one.

Can you do better?

It is very difficult to come up with the solution from scratch (but do try!). Here are a few numbers already filled in, if you would like a starting point (and as a pdf).

The solution is not unique, but here is Ludgate’s solution (and as a pdf).

In a Latin Square each symbol or colour occurs exactly once in each row and exactly once in each column.

Can you complete this one? It is from Brilliant.org where you can do it interactively.

Now for the hard challenge …

Thanks to NRich for this great puzzle

Task 1  – Rectangles

Today we were inspired by this amazing video from 3 Blue 1 Brown.

Can you always draw a rectangle on any shape? How about a square?

Here is a series of challenges on paper and geogebra to explore this fascinating topic.

Here are the 1st and 2nd explorations we did today to learn about modular arithmetic and a fantastic geogebra app.

From Wikipedia:

“The 100 prisoners problem is a mathematical problem in probability theory and combinatorics. In this problem, 100 numbered prisoners must find their own numbers in one of 100 drawers in order to survive. The rules state that each prisoner may open only 50 drawers and cannot communicate with other prisoners after the first prisoner enters to look in the drawers. At first glance, the situation appears hopeless, but a clever strategy offers the prisoners a realistic chance of survival.

Anna Gál and Peter Bro Miltersen first proposed the problem in 2003.”

In maths club we tried this problem with just 10 prisoners. You can read about the strategy here.

We analysed the percentage chance of winning by looking at the different possible cycles for 10 prisoners.

We worked on a problem presented in Brooklyn 99:

There are twelve identical looking islanders and a seesaw. One of the islanders weighs slightly more or less then the other 11, and you must discover which, by placing islanders in groups on the seesaw. However only three measurements are allowed.

You can test your method with this app:

We found this activity on Nrich and it was originally in one of Brian Bolt’s books and developed by MEDIAN in their collection of interesting number resources.

This Daisy is special because you can make every number from 1 to 25.
You are only allowed to add neighbours (numbers touching each other) and you can
only use each number once in a sum.

1 = 1
2 = 2
3 = 1 + 2
4 = 4
5 = 5
6 = 2 + 4
7 = 1 + 2 + 4
8 = 5 + 1 + 2
9 = 4 + 5
10 = 7 + 1 + 2
11 = 5 + 4 + 2
12 = 5 + 4 + 2 + 1
13 = 7 + 6
14 = 7 + 6 + 1
15 = 7 + 6 + 2
16 = 7 + 6 + 2 + 1
17 = 7 + 5 + 4 + 1
18 = 7 + 5 + 4 + 2
19 = 7 + 5 + 4 + 2 + 1
20 = 5 + 7 + 6 + 2
21 = 5 + 7 + 6 + 2 + 1
22 = 4 + 5 + 7 + 6
23 = 5 + 7 + 6 + 1 + 4
24 = 5 + 7 + 6 + 2 + 4
25 = 5 + 7 + 6 + 2 + 4 + 1

Can you do better than this with a different set of numbers?
The challenge is to find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25.

You can check your ideas here:

Could you adapt the program above to search for a higher solution?

To write the program above we needed to find all the combinations from a general daisy like this:

For example you can have a+b or a+f but not a+d. Can you find all the combinations? There is a worksheet to fill in here (with the solution on the second page). An editable version is here.