Problems

Arne has a cube which is pink on top and orange on bottom, yellow on right and white on left, turquoise on front and dark blue at the back. He rotates this once so that it looks different. Could he perform the same rotation four more times and get back to the original colouring?

Sam the magician shuffles his hand of six cards: joker, ace (\(A\)), ten, jack (\(J\)), queen (\(Q\)) and king (\(K\)). After his shuffle, the relative order of joker, \(A\) and \(10\) is now \(A\), \(10\), joker. Also, the relative order of \(J\), \(Q\) and \(K\) is now \(Q\), \(K\) and \(J\).
For example, he could have \(A\), \(Q\), \(10\), joker, \(K\), \(J\) - but not \(A\), \(Q\), \(10\), joker, \(J\), \(K\).

How many choices does Sam has for his shuffle?

The perimeter of the triangle \(\triangle ABC\) is \(10\). Let \(D,E,F\) be the midpoints of the segments \(AB,BC,AC\) respectively. What is the perimeter of \(\triangle DEF\)?

Let \(\triangle ABC\) be a triangle and \(D\) be a point on the edge \(BC\) so that the segment \(AD\) bisects the angle \(\angle BAC\). Show that \(\frac{|AB|}{|BD|}=\frac{|AC|}{|CD|}\).

Show that if \(1+2+\dots+n = \frac{n(n+1)}{2}\), then \(1+2+\dots+(n+1) = \frac{(n+1)((n+1)+1)}{2}\).

Show that \(1+2+\dots+n = \frac{n(n+1)}{2}\) for every natural number \(n\).

Show that if \(1+2^1+2^2+\dots+2^{10} = 2^{11} - 1\), then \(1+2^1+2^2+\dots+2^{11} = 2^{12} - 1\).

Show that \(1+2^1+2^2+\dots+2^n = 2^{n+1} - 1\) for every natural number \(n\).

What is wrong with the following proof that “all rulers have the same length" using induction?

Base case: suppose that we have one ruler, then clearly it clearly has the same length as itself.

Assume that any \(n\) rulers have the same length for the induction hypothesis. If we have \(n+1\) rulers, the first \(n\) ruler have the same length by the induction hypothesis, and the last \(n\) rulers have the same length also by induction hypothesis. The last ruler has the same length as the middle \(n-1\) rulers, so it also has the same length as the first ruler. This means all \(n+1\) rulers have the same length.

By the principle of mathematical induction, all rulers have the same length.

Given a series of statements enumerated by the natural numbers, the strong induction principle says the following. Suppose that

• The 1st statement is true (the base case).

• Whenever all statements up to and including the \(n\)th statement is true, the \((n+1)\)th statement is also true (induction step).

Then the statement is true for all natural numbers. Show that the strong induction principle works.