Lycee Francais SAMI Maths Club

The aim of this puzzle is to switch the red frogs with the blue frogs according to the following rules:

• The left set of frogs can only move right, the right set of frogs can only move left.
• Frogs can move forward one space, or move two spaces by jumping over another counter.
• The puzzle is solved when the two sets of frogs have switched positions.

How many moves does this puzzle take?

How about if we change the number of frogs? Can you find a rule for the number of moves it will take based on how many frogs there are?

NRich has an applet with nice graphics here, but you are allowed to go backwards in their version, so it is is best to use this one to help you count the moves.

For interested facilitators extra notes and the solutions are here.

Today we solved an intriguing homework riddle with the help of a Venn diagram.  We had clues such as

“Seven students in total have Maths homework but Miya isn’t one of them.”

and

“Rory has more homework than Aidan.”

Using all the clues, which you can find here, we were able to work out which students had which pieces of homework.

It was a nice challenge! The solution can be found here. This activity was originally found on the TES website.

We spent this maths club playing the highly addictive game Square it from Nrich.

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.

… will the 28th of July 2061 be?  (Next predicted sighting of Halley’s comet)

… was the 4th of April 1965? (Robert Downey Jr ‘s birth)

… was the 17th of July 1789? (Prise de la Bastille)

We puzzled over these two questions having to think about where to start, the number of days in each month, and the rules for leap years.

It was fun to try different strategies by hand, but then we used the mod function and the floor function in Zeller’s algorithm to find quickly the day of the week for any date.

Today we explored one of the most famous recursive sequences called The Logistic Map.

Try this activity leading on to exploring the sequence on Geogebra. There are Geogebra instructions here if you get stuck.

We were inspired for this activity by this amazing video from Ben Sparks on Numberphile. It explains how this activity links to one of the Feigenbaum constants, pseudo ramdom numbers and Chaos Theory. Definitely worth watching!

The first question of today is how many sides a piece of paper has. So the answer is 2 of course. Now we want to find a way of folding the paper in such a way that there is only one side. One way to verify this is to take a pen and colour the folded piece of paper. If you do not need to turn the piece of paper around, there is only one side. The method to do this is to twist the piece of paper by half a turn and you will obtain a möbius band.

This is all explained in this presentation from Suffolk maths, that also contains a brilliant exploration table. Your challenge is to fill in the first three lines and to come up with three other results to add to the table below.

Good luck!

This week we worked through another brilliant puzzle from the Math Pickle team called McGuire the Gathering.

Lots of fun trying to solve both on pen and paper and with excel!

We used this website to solve a series of challenges where you start with front, side and top elevations to build the full picture of 3D shape – building up to using only two views and silhouettes.

In pairs we played the following game a couple of times.

One person is heads (player 1), the other person is tails (player 2). Toss the coin a maximum of 10 times, player 1 gets a point if the coin lands on heads, and player 2 gets a point if the coin lands on tails. If one person becomes three points ahead of the other the game stops and they win. If no one is 3 points ahead after ten tosses the game is a tie.

Which of the following scores are impossible in this game and why? 5-2, 7-4, 6-2

Think of ways the game could end in a tie.

What is the chance of winning this game?

This following table will help.

Fill in the table shading out impossible final scores, highlighting winning final scores for Player 1 and Player 2, and highlighting final scores which indicate a draw.
Then write in each box how many ways there are to end up at each score.

A few boxes have been filled in – for example 3-0 is a winning score for player 1, and there is only 1 way to get to that score (1-0 then 2-0 then 3-0). There are two ways to get to a score of 1-1 (0-1 then 1-1 or 1-0 then 1-1). There is a 1 in the 0-0 box as there is just one way to start the game.

See  here for the solution, contained in an excellent presentation by Alex in the Further Maths class in PAL. This puzzle is the starting point for analysing how likely you are to win a tennis match if you only have a one third chance of winning each point. The idea and diagrams in the presentation were taken from the book Game, Set and Math by Ian Stewart.