Take a number e.g. 537
Multiply its digits 5 x 3 x 7 = 105
Repeat with 105’s digits. 1 x 0 x 5 = 0
Stop when you reach a single digit.
So 537 took 2 steps.
Can you find a number that takes more steps? You can test your numbers here:
Then can you find the smallest number that takes 1,2,3,4… steps? Click run when you want to see the answers.
Thanks to numberphile for this problem – their video is here.
Let’s say we assign each letter of the alphabet a different number.
The value of a word is the value of the sum of its letters.
For example:
A = 1
B = 2
C = 3, etc.
What is the value of the word FOUR? Answer = 6 + 15 + 21 + 18 = 60
A number is “nice” if its value is equal to itself.
Now, you can use any assignment of letters to integers you like (provided each letter is assigned to a distinct integer).
Can you find an assignment of letters for which all of the values from 1 to n are “nice”?
What is the largest value of n? Answer here.
We found this solution for the assignment of the letters.
In a swimming relay race there are four team members and each one swims just one of the strokes: butterfly, backstroke, breaststroke or front-crawl.
You are the swimming coach for the Hungarian team in the Olympics, who should you choose for each stroke based on their current times in seconds?
| Butterfly | Backstroke | Breaststroke | Front-crawl | |
| Adam | 35 | 36 | 36 | 25 |
| Boris | 31 | 31 | 35 | 23 |
| Carl | 33 | 35 | 39 | 26 |
| Dean | 32 | 38 | 33 | 24 |
Remember it is the fastest team that wins!
Once you have tried this yourself you could look here at an algorithm that can be used, hints to try it on excel and the solution.
This one is a bit trickier …
| Butterfly | Backstroke | Breaststroke | Front crawl | |
| Ella | 29 | 34 | 31 | 24 |
| Freya | 32 | 28 | 30 | 25 |
| Grace | 30 | 33 | 27 | 26 |
| Holly | 31 | 29 | 32 | 23 |
We had fun in maths club trying to estimate the number of popcorn kernels in this jar. We thought about optimal stacking, averaging guesses, lines of best fit and finally used a smaller sized container to refine our estimate.
You can see our thought processes here.
Click on the link below to try the fun puzzle above – can you turn all the lights out for any size grid?
Square numbers
Triangular numbers
Square Triangular Numbers
1 appears on both lists so is a square triangular number.
0 could also be described as a square number (0x0) and is technically the first triangular number so 0 is also a square triangular number.
Can you find any more square triangular numbers (numbers that appear on both lists)?
Solutions, notes and investigations are here.
Today we played a game where you have to take card and try and make 15 before your opponent. It relates to a game you have almost certainly played before. You could use playing cards to play and then discover the secret.
Here is a worksheet with all the rules and extensions to a hexagon puzzle, and the solutions to the hexagon are here.
We played a game posted on youtube by Mr Allen and extended it to different bases.
Set up
2-4 players. Using Ace -10 cards only.
Rules:
7 (or 6 if there are four of you) cards are placed face up for each player and two in the middle. The player who gets rid of their cards first wins.
You can put a card in the middle (on top of either pile) when one of your cards is an answer to an expression using the two cards in the middle. eg. if 6 and 2 are in the middle possible cards that can be played are: 8 (6+2) or 4 (6-2) or 3(6 2) or 2(6×2 – 12 is the answer but we use the last digit only).
Each player has to say the expression as they put the card down. eg 6 twos are 12 – placing the 2 or 6 minus 2 is 4. All players play on until no cards can be played. Then the dealer deals each player a card then put one on one of the two middle piles. Players then race to place cards until someone gets rid of all their cards!
Tasks
Play a few games of this and try some variations:
In the original game we used the last digit if the calculation was bigger than 10. This is equivalent to subtracting 10 until you have an answer that is the right size. This is called Mod 10. Mod 8 means that if you do e.g. 7 + 3 = 10 and get an answer that is too big you subtract 8 until it is the right size. 10 – 8 = 2.
More examples: 5 + 6 = 11 = 3
6 × 2 = 12 = 4
What is the best version (10,8 or 7) to play in terms of chances to play a card? You could use this worksheet to help you decide.
Task 1 – Warm up
To see how to draw circles, plot points and pick random numbers have a look at this code and adapt it to draw a face:
Task 2
Find the area of the square below
Find the area of the circle
Find the answer to (4 x area of circle / area of square)
Task 3
Adapt the app below to estimate pi. You are aiming for this:
Note that the formula for a circle is x**2+y**2=r**2
where ** on python is squared.
Suppose there was one of six toy animals inside your favorite box of cereal and you would like to collect them all.
You could be lucky and only buy six boxes, but how many boxes of cereal would you expect to have to buy on average?
With thanks to this website we carried out an experiment with dice to simulate this problem. We rolled a dice until we had thrown all the numbers 1 to 6 and recorded how many rolls it took.
Click here for a pdf with several blank tables to cut out and use.
We then used python to generate 1000 trials and compared our average to the theoretical average we calculated. Can you try and program this in the trinket below?
Or, for very advanced mathematicians, you could try and find the theoretical probability. Worksheet here.
Solution in python is here, and the theoretical probability can be found in this article.
I am 35 years old.
My friend is 53 years old.
35 and 53 are mirror birthdays.
We also looked at this puzzle in binary!
Inspired by this fantastic numberphile video with Ben Sparks we are explored sequences of aliquot sums. An aliquot sum is the sum of proper factors of numbers.
The number 15 has factors 1, 3, 5, 15 but the proper factors don’t include the number itself so 15 has proper factors 1, 3, 5. Adding these together gives an aliqout sum of 1+3+5 = 8.
Now we find the aliquot sum of 8, which is 1+2+4 = 7. Then find the aliquot sum of 7, which is just 1 as 7 is a prime number. If we get to 1 we stop.
So 15 – 8 – 7 – 1.
Try with different starting numbers, do all numbers go to 1?
Aliquot 1 – Program to find the aliquot sum of a number
Aliquot 2 – Program to find the aliquot sequence of a starting number
Aliquot 1
Use the computer to find the sum of the factors of 220.
What is the sum of the factors of the answer?
Which one of the following pairs of numbers behaves in the same way?
607, 917
624, 1112
1184, 1210
1554, 2094
Aliquot 2
Can you find a sequence that goes up and then down?
Can you find a sequence that goes up and then down multiple times? Try 138 …
The number 276 is special as it is the smallest number that we don’t know whether or not it ends up at 1. This is an unsolved mathematics problem.
Today we worked on challenging fraction questions sticking to ancient Egyptian methods – only unit fractions and you can’t repeat a fraction.
For example, can the following be expressed as a sum of distinct unit fractions?
This led to the greedy algorithm and some programming on excel:
See here for all the details and solutions.
We are counting in whole numbers always starting at the number 1.
1,2,3,4, …
Schur numbers tell you the highest number that you can count to using k different colours before you’re forced to have an all same-coloured solution to a + b = c.
Example
Is this a valid colouring for k = 3 (3 colours)?
No, because 2 + 2 = 4, 1 + 5 = 6 and 1 + 6 = 7 and same coloured sums are not allowed.
Challenge
For 1 colour, let’s say red, we can only count up to the number 1 like this:
1
For 2 colours, let’s say red and blue, we can count up to 4 like this:
1 2 3 4
Can you explain why we can’t add the number 5?
What is the highest number you can count to using 3 colours?
You can check solutions here:
Here is a python program that generates all the solutions:
https://www.programiz.com/online-compiler/2S5O1C7J96Ys7
See here for the printable puzzle with further explanations and here for a video including the solutions (that are known so far …).
University of Southampton are running a National Cipher Challenge.
Introduction to some of the ciphers that will be used is here.
Competition is https://www.cipherchallenge.org/
There is a great set of tools to use:
https://www.cipherchallenge.org/tools/
Please sign up for the challenge!
Contact Mrs Fleming on [email protected] if you need the monitoring pin to complete the registration.
Here is a series of puzzles with red and blue hats that will challenge your logical skills.
Today we did one of our favourite puzzles inspired by this Numberphile video by Neil Sloane.
Starting with 2 ones, what is the maximum number you can place on an infinite grid according to the rule:
“You can place a number if all the numbers immediately around it add up to itself”
Here is a way to get up to 4:
But then we are stuck as there is nowhere to put the 5. Can you do better than 4?
You can choose where to place the 2 ones to start with, for example:
This puzzle can be tried with any number of 1s to start with.
What is the maximum number you can place starting with three 1s?
Some solutions are here.
This investigation is based on the ancient Greek Apollonius Problem: Can you draw a circle that just touches three given circles? How many can you draw?
We tried this by hand and then on Geogebra following these instructions.
The maximum number of circles possible was unsolved for more than 2000 years, but we solved it! For the solutions and more see the links below:
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
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:
One solution is discussed here, and then another more elegant solution is here.
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.
