Problems

In this example we will discuss division with remainder. For polynomials \(f(x)\) and \(g(x)\) with \(\deg(f)\geq \deg(g)\) there always exists polynomials \(q(x)\) and \(r(x)\) such that \[f(x)=q(x)g(x)+r(x)\] and \(\deg(r)<\deg(g)\) or \(r(x)=0\). This should look very much like usual division of numbers, and just like in that case, we call \(f(x)\) the dividend, \(g(x)\) the divisor, \(q(x)\) the quotient, and \(r(x)\) the remainder. If \(r(x)=0\), we say that \(g(x)\) divides \(f(x)\), and we may write \(g(x)\mid f(x)\). Let \(f(x)=x^7-1\) and \(g(x)=x^3+x+1\). Is \(f(x)\) divisible by \(g(x)\)?

We start with the point \(S=(1,3)\) of the plane. We generate a sequence of points with coordinates \((x_n,y_n)\) with the following rule: \[x_0=1,y_0=3\qquad x_{n+1}=\frac{x_n+y_n}{2}\qquad y_{n+1}=\frac{2x_ny_n}{x_n+y_n}\] Is the point \((3,2)\) in the sequence?

Consider a graph with four vertices and where each vertex is connected to every other one (this is called the complete graph of four vertices, sometimes written as \(K_4\)). We write the numbers \(10,20,30,\) and \(40\) on the vertices. We play the following game: choose any vertex, and subtract three from that vertex, and add one to each of the three other vertices, so an example could be:

After playing this game for some number of steps, can we make the graph have the number \(25\) on each vertex?

Every year the citizens of the planet “Lotsofteeth" enter a contest in which they find the person with the most teeth. The judge notices that no one this year is toothless and that there are more people than the number of teeth in any single person. Is it true that there are two people with exactly the same number of teeth and why?