Problems

Let’s play some games today! We will play a classic game known as nim, which is thought to be one of the oldest games.

Typically people play nim using matchsticks, though stones and coins are popular too. There are a few heaps of matchsticks in nim. Players take turns to remove matchsticks from a heap of their choosing. The player can remove any number of matchsticks they wish from that heap. Whoever has no matchsticks left to take loses.

This following position will be written as \(\text{Nim}(3,3,3)\):

As another example, this is \(\text{Nim}(1,2,3,4)\):

We will omit heaps of size zero, so \(\text{Nim}(3,0,3,0,3)\) is the same as \(\text{Nim}(3,3,3)\).

Nim is important because a large class of games are equivalent to it despite its simple appearance. The interested reader should look up “Sprague-Grundy Theorem".

Let us introduce a few terms that will be helpful for analyzing games. A game \(G\) consists of some positions and a set of rules. A position \(g\) in the game \(G\) is called a winning position if the player starting this turn has a winning strategy. This means as long as the player starting this turn continues to play optimally, the second player has to lose. Conversely, a position \(g\) is a losing position if the player starting this turn has no winning strategy.

For any real number \(x\), the absolute value of \(x\), written \(\left| x \right|\), is defined to be \(x\) if \(x>0\) and \(-x\) if \(x \leq 0\). What are \(\left| 3 \right|\), \(\left| -4.3 \right|\) and \(\left| 0 \right|\)?

Let \(x\) and \(y\) be real numbers. Prove that \(x \leq \left| x \right|\) and \(0 \leq \left| x \right|\). Then prove that the following inequality holds \(\left| x+y \right| \leq \left| x \right|+\left| y \right|\).

There are \(n\) balls labelled 1 to \(n\). If there are \(m\) boxes labelled 1 to \(m\) containing the \(n\) balls, a legal position is one in which the box containing the ball \(i\) has number at most the number on the box containing the ball \(i+1\), for every \(i\).

There are two types of legal moves: 1. Add a new empty box labelled \(m+1\) and pick a box from box 1 to \(m+1\), say the box \(j\). Move the balls in each box with (box) number at least \(j\) up by one box. 2. Pick a box \(j\), shift the balls in the boxes with (box) number strictly greater than \(j\) down by one box. Then remove the now empty box \(m\).

Prove it is possible to go from an initial position with \(n\) boxes with the ball \(i\) in the box \(i\) to any legal position with \(m\) boxes within \(n+m\) legal moves.

Given a natural number \(n\), find a formula for the number of \(k\) less than \(n\) such that \(k\) is coprime to \(n\). Prove that the formula works.

Scrooge McDuck has \(100\) golden coins on his office table. He wants to distribute them into \(10\) piles so that no two piles contain the same amount of coins. Moreover, no matter how you divide any of the piles into two smaller piles, among the resulting \(11\) piles there will be two with the same amount of coins. Find an example of how he could do that.

Today we will draw lots of pictures.

The subject is Topology. It is often called “rubber-sheet geometry" because while it is the study of shapes, topologists typically do not pay too much attention to rigid notions like angle and lengths. We have much more flexibility in topology. Some common words describing the operations here might include “gluing", “stretching", “twisting" and “inflating".

Although we will not define continuity, it is a more or less intuitive idea. Topological operations should be continuous. If you have a line segment, no amount of stretching, twisting or bending can make it into two disconnected segments.

Let \(A=\{1,2,3\}\) and \(B=\{2,4\}\) be two sets containing natural numbers. Find the sets: \(A\cup B\), \(A\cap B\), \(A-B\), \(B-A\).

Let \(A=\{1,2,3,4,5\}\) and \(B=\{2,4,5,7\}\) be two sets containing natural numbers. Find the sets: \(A\cup B\), \(A\cap B\), \(A-B\), \(B-A\).

Given three sets \(A,B,C\). Prove that if we take a union \(A\cup B\) and intersect it with the set \(C\), we will get the same set as if we took a union of \(A\cap C\) and \(B\cap C\). Essentially, prove that \((A\cup B)\cap C = (A\cap C)\cup (B\cap C)\).

\(A,B\) and \(C\) are three sets. Prove that if we take an intersection \(A\cap B\) and unite it with the set \(C\), we will get the same set as if we took an intersection of two unions \(A\cup C\) and \(B\cup C\). Essentially, prove that \((A\cap B)\cup C = (A\cup C)\cap (B\cup C)\). Draw a Venn diagram for the set \((A\cap B)\cup C\).