Problems

Let \(n\) be a positive integer. Prove that it’s impossible to have a closed knight’s tour on a \(4\times n\) grid.

Find an open knight’s tour on a \(2\times2\times2\) cube.

Four football teams play in a tournament. There’s the Ulams (\(U\)), the Vandermondes (\(V\)), the Wittgensteins (\(W\)) and the Xenos (\(X\)). Each team plays every other team exactly once, and matches can end in a draw.
If a game ends in a draw, then both teams get \(1\) point. Otherwise, the winning team gets \(3\) points and the losing team gets \(0\) points. At the end of the tournament, the teams have the following points totals: \(U\) has \(7\), \(V\) has \(4\), \(W\) has \(3\) and \(X\) has \(2\).

Work out the results of each match, including showing that there’s no other way the results could have played out.

Prove by induction that \(n^{n+1}>(n+1)^n\) for integers \(n\ge3\).

What is the following as a decimal? \[\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}.\]

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.

Let \(n\) be a positive integer. Show that \(1+3+3^2+...+3^{n-1}+3^n=\frac{3^{n+1}-1}{2}\). You may like to use induction.

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.