Lycee Francais SAMI Maths Club

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.

Take an A4 piece of paper.

Fold it a few times.

Then punch a hole through the paper.

Before you unfold your paper, can you draw on a fresh piece of paper where you think the holes will be?

Then can you fold a piece of paper and punch it to make the holes look like this:

Then can you fold and cut a piece of paper to make it wider then it was originally? Can you cut in such a way that you can step through it?

Similar ideas can be found here. Solution to stepping through it here.

On dotted paper we explored circles of different radius, counting the dots on or inside the circle. Brilliantly the number of dots is very close to the area of the circle.

We wrote this program in Python to check the ratio of dots to area for different size circles.

Challenge 1
Imagine you have an endless supply of 5p and 7p coins. You could make exactly 20p with four 5p coins. Or you could make exactly 19p with two 7p coins and a 5p coin.

What is the biggest amount that you cannot make? Can you explain why you can make all amounts after this one?

Challenge 2
The biggest number you cannot make given coins of value x and y is called the Frobenius number.

Can you think of a pair of numbers that wouldn’t have a Frobenius number?

What must be true about the two numbers for them to have a Frobenius number?

Challenge 3
Can you come up with a formula for the Frobenius number, if it exists, for two numbers x and y?

Solutions here.

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

We worked out the magic/maths behind card tricks 1 and 4 from this set of tricks.

In the 501 game of darts players take turns at throwing 3 darts to reduce their score to zero.

A “checkout” refers to the process of finishing a game by reducing a player’s score to exactly zero, by hitting a double or the bullseye (50 points) with the final dart.

For example, if a player has 40 remaining, they can hit the double 20 (D20) to win.

Here is an example of a 3 dart checkout:

120 : T20 20 D20    (treble 20, single 20, double 20)

1. The maximum checkout is 170. How can you make this?
2. For which numbers between 140 to 170 can you find a three dart checkout?

140		150		160
141		151		161
142		152		162
143		153		163
144		154		164
145		155		165
146		156		166
147		157		167
148		158		168
149		159		169

Most darts players like to aim to finish on D20, D18, D16 or maybe Bullseye.

1. A player has 94 left with three darts. They aim for T18
2. What is their checkout if they hit T18?
3. What is their next dart if they hit a single 18 instead?

2. Imagine you have three darts. Would you rather a score of 32 or 30 left? Think about what happens if you just miss D16 and get single 16 left versus just missing D15 and getting single 15.

3. What checkouts can you find for three darts on 107? What would be the best option do you think?

Extra reading

Logic behind checkouts:

Checkouts Explained

Checkout game:

Darts Math Quiz application