Maths Circles
Maths Circles are free maths-loving sessions for secondary school pupils. They encourage and develop structured thinking and problem-solving skills, and are aimed at cultivating children’s abilities to articulate ideas and construct rigorous proofs.
It is a social club for maths loving kids – a place where everyone shares a passion for puzzle solving and mathematics.
Maths Circles are held weekly, each session is 1.5 hours long. They are best suited for students in school years 7–11. No prior knowledge is expected from the students.
The Maths Circles programme is not a course with a beginning and an end. It is an ongoing social activity that can be attended for many years to come.
Contents
Curriculum
Our curriculum represents an innovative adaptation of the well established Maths Circles programme, designed to foster problem-solving abilities and cultivate rigorous scientific reasoning.
The program encompasses an extensive array of subjects spanning Algebra, Calculus, Combinatorics, Game Theory, Euclidean Geometry, and Graph Theory.
These topics lie outside the typical school syllabus and generally maintain independence from one another. It makes Maths Circles accessible to students from diverse educational backgrounds and permits kids to easily rejoin even if they miss several sessions.
Tutors
Maths Circles are led by postgraduate and PhD students of Mathematics of the best Universities of the UK. They are trained and supervised by the We Solve Problems staff.
Being much closer in age to the pupils, the tutors not only help them through advanced maths, but also serve as role models of ‘real life mathematicians’.
Want to join our tutors’ team? Fill in this form.
How to Attend
Maths Circles are free of charge for all students of the school years 7 to 11, but depending on the location we can be experiencing very high demand. Participation in the London Verbal Maths Challenge is a prerequisite for attending Maths Circles at King’s College, University College, and Queen Mary University of London, as well as at Keble College (University of Oxford).
To check currently open Circles, please consult the section below.
To register your interest in attending one of the Circles, please proceed below to the Circle of your interest and fill there the pre-registration form.
Circles in 2023/24
Maths Circles are run throughout the school year, from September to May. We have all the usual breaks for half terms and holidays.
Please note that, depending on the location we might be experiencing very high demand.
Bath
Cardiff
London
Oxford
Warwick
Session Example
Each session starts with an introduction of a mathematical concept by one of the senior tutors, and analysing and discussing solutions to several examples.
Then, the children are asked to try to solve a given set of problems crafted around the topic of the day.
The students are encouraged to interact with the tutors, either to present their solutions or to ask for hints. A student can raise a hand at any moment, and a tutor will come by to discuss the problem and its solution onу-to-one with the student.
Have a look at this PDF file to see an example of a real life Problem sheet.
Sample Problems
Problem 1
Two goblins Krok and Grok, are playing a game with a pile of gold. Each goblin can take any positive number of coins no larger than 9 from the pile. They make moves one after another. There are 3,333 coins in total, the goblin who takes the last coin wins. Who will win, if Krok goes first?
See Solution
We could analyse exactly which positions are winning and which are losing – but instead we can how one always makes 10 from two numbers less than, or equal to 10. All the positions divisible by 10 are losing, and all the other positions are winning. If one player is starting from a number divisible by 10, they say any number x and then the other player says the number 10−x, and after two moves the number has been reduced by 10. That means we just need to check if 3,333 is divisible by 10… but of course its not. That means Krok is in the winning position, he needs to just take 3 and then match Groks’s number of coins to 10 and he will be the person to have the last move.
Problem 2
How many 7-digit numbers, larger than 6,000,000, are there such that the product of their digits equals 42?
See Solution
Consider a number S in the set of first 1,000 natural numbers, which can be expressed as the sum of 6 consecutive integers starting with some integer m-2: S = m – 2 + m – 1+m+m+1+ m+2 + m+3 = 6m +3. Therefore if S can be expressed as the sum of 6 consecutive integers then S has remainder 3 modulo 6. Conversely, if S has remainder 3 after division by 6, then we automatically find m and therefore the expression for S as the sum of 6 consecutive integers. Any number that isn’t of this form cannot expressed like this. We just need to find how many such numbers are in this set.
The largest such number in the set {1,2,.. 1,000} can be found by dividing 1,000 by 6. The result comes up as 166 +2/3, that means 1,000 = 6 * 166 +4. That means 999=6*166+3. That is the largest working number from this set, with m=166, the smallest number is 3 with m=0, so there are 167 such numbers in total.
Problem 3
Show that if numbers a-b and c-d are divisible by 11, then ac-bd and ad-bc are also both divisible by 11.
See Solution
Assume a has remainder r when divided by 11 and b has remainder s. Then, the remainder of a-b is a remainder of r-s. But we also know that the remainder of a-b is zero since this number is divisible by 11. That means, r = s, because r and s have to be smaller than 11 and there is no way for r-s to be divisible by 11 in that case, if they aren’t equal.
Thus we know that a and b have the same remainders in division by 11, and we can prove the same statement about c and d. We need to show that ac has the same remainder as bd and that ad has the same remainder as bc.
Let’s assume that the numbers c and d have the remainder t when divided by 11. Then, the remainder of ac is the remainder of rt and the remainder of bd is also the remainder of rt. The same is true for remainders of ad and bc, they are all the same as the remainder of rt. Thus, both ac-bd and ad-bc are divisible by 11.
There is another solution that doesn’t use remainders explicitly:
ac -bd = ac – bc + bc – bd = c * (a-b) – b * (c -d)
It is a sum of two numbers, both of which are divisible by 11, so it must be divisible by 11.
Actually, this is how you prove that the remainder of a product is a remainder of the product of remainders. It is enough to say b = r, the remainder of a, and d = t, the remainder of c. Then clearly a-r and c-t are both divisible by 11 or whatever we are dividing by, and that means ac-rt is divisible by the same number, so the remainder of ac is the same as the remainder of rt.
You can find a full scale collection of such problems in our new project Maths Problems Database.