Problems

Prove that \(3\) always divides \(2^{2n}-1\), where \(n\) is a positive integer.

Adi and Maxim play a game. There are \(100\) sweets in a bowl, and they each take in turns to take either \(2\), \(3\) or \(4\) sweets. Whoever cannot take any more sweets (since the bowl is empty, or there’s only \(1\) left) loses.

Maxim goes first - who has the winning strategy?

Michelle and Mondo play the following game, with Michelle going first. They start with a regular polygon, and take it in turns to move. A move is to pick two non-adjacent points in one polygon, connect them, and split that polygon into two new polygons. A player wins if their opponent cannot move - which happens if there are only triangles left. See the diagram below for an example game with a pentagon. Prove that Michelle has the winning strategy if they start with a decagon (\(10\)-sided polygon).

Let \(n\) be a positive integer. Show that \(1+3+3^2+...+3^{n-1}+3^n=\frac{3^{n+1}-1}{2}\).

Show that all integers greater than or equal to \(8\) can be written as a sum of some \(3\)s and \(5\)s. e.g. \(11=3+3+5\). Note that there’s no way to write \(7\) in such a way.

You may have seen the first example (seen below) previously. Before we get to that, let’s introduce some notation. We write \(K_n\) for the complete graph on \(n\) vertices - that is, every possible edge is present.

Note that edges don’t have a direction, and are between pairs of different vertices. Ramsey theory looks at what happens when we colour every edge in \(K_n\) either red or blue. Are we guaranteed a red \(K_3\) subgraph? No, because we could just colour every edge in \(K_n\) blue. Instead, we ask if we are guaranteed a red \(K_3\) or a blue \(K_3\) subgraph? It turns out yes, so long as \(n\) is big enough.

We can then look at extensions to this problem. We write \(R(s,t)\) for the least number \(n\) such that whenever you colour the edges of \(K_n\) in red or blue, then you’re guaranteed a red \(K_s\) or a blue \(K_t\). By least \(n\), this means that there’s a colouring of the edges of \(K_{n-1}\) with no red \(K_s\) or blue \(K_t\).

You may like to use the inequality \(R(m,n)\le R(m,n-1)+R(m-1,n)\). Furthermore, when both \(R(m,n-1)\) and \(R(m-1,n)\) are even, we have the stronger inequality of \(R(m,n)\le R(m,n-1)+R(m-1,n)-1\).

Explain why you can’t rotate the sides on a normal Rubik’s cube to get to the following picture (with no removing stickers, painting, or other cheating allowed).

A circle with centre \(A\) has the point \(B\) on its circumference. A smaller circle is drawn inside this with \(AB\) as a diameter and \(C\) as its centre. A point \(D\) (which is not \(B\)) is chosen on the circumference of the bigger circle, and the line \(BD\) is drawn. \(E\) is the point where the line \(BD\) intersects the smaller circle.

Show that \(|BE|=|DE|\).

Are there any two-digit numbers which are the product of their digits?

The sum of Matt’s and Parker’s ages is \(63\) years. Matt is twice as old as Parker was when Matt was as old as Parker is now. How old are they? (Show that there’s no other ages that they could have)