Problems

In the following grid, how many different ways are there of getting from the bottom left triangle to the bottom right triangle? You must only go from between triangles that share an edge and you can visit each triangle at most once. (You don’t have to visit all of the triangles.)

You and I are going to play a game. We have one million grains of sand in a bag. We take it in turns to remove \(2\), \(3\) or \(5\) grains of sand from the bag. The first person that cannot make a move loses.

Would you go first?

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.

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.

A parliament has 650 members. In this parliament there is only one house and every member has at most three enemies. We wish to split this parliament into two separate houses in such a way that each member will have at most one enemy in the same house as them. We assume that hard feelings among members of parliament are mutual, namely if \(A\) recognises \(B\) as their enemy, then \(B\) also recognises \(A\) as their enemy.

Is this splitting possible?

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 \((A\cap B)\cup C\).

Let \(A,B\) and \(C\) be three sets. Prove that if we take an intersection \(A\cap B\) and intersect it with the set \(C\), we will get the same set as if we took an intersection of \(A\) with \(B\cap C\). Essentially, prove that it does not matter where to put the brackets in \((A\cap B)\cap C = A\cap (B\cap C)\). Draw a Venn diagram for the set \(A\cap B\cap C\).
Prove the same for the union \((A\cup B)\cup C = A\cup (B\cup C) = A\cup B\cup C\).