Eleven sages were blindfolded and on everyone’s head a cap of one of \(1000\) colours was put. After that their eyes were untied and everyone could see all the caps except for their own. Then at the same time everyone shows the others one of the two cards – white or black. After that, everyone must simultaneously name the colour of their caps. Will they succeed?
The recertification of the Council of Sages takes place as follows: the king arranges them in a column one by one and puts on a cap of white or black colours for each. All the sages see the colours of all the caps of the sages standing in front, but they do not see the colour of their own and all those standing behind. Once a minute one of the wise men must shout one of the two colours. (each sage shouts out a colour once). After the end of this process the king executes every sage who shouts a colour different from the colour of his cap. On the eve of the recertification all one hundred members of the Council of Sages agreed and figured out how to minimize the number of those executed. How many of them are guaranteed to avoid execution?
At the elections to the High Government every voter who comes, votes for himself (if he is a candidate) and for those candidates who are his friends. The forecast of the media service of the mayor’s office is considered good if it correctly predicts the number of votes at least for one of the candidates, and not considered good otherwise. Prove that for any forecast voters can show up at the elections in such a way that this forecast will not be considered good.
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 \(g(x)\)?
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 \(g(x)\)?
In this example we discuss the Factor Theorem. First, let us recall the following concept: if \(P(x)\) is a polynomial, then a number \(z\), is a root of \(P(x)\) if \(P(z)=0\). For example, \(x=1\) is a root of the polynomial \(Q(x)=x-1\). Show:
If \(0\) is a root of \(P(x)\), i.e: \(P(0)=0\), then \(P(x)=xQ(x)\) for some polynomial \(Q(x)\).
Use part 1 to show the Factor Theorem: if \(z\) is a root of \(P(x)\), then \(P(x)=(x-z)K(x)\) for some polynomial \(K(x)\).
In this example we discuss one of Vieta’s formulae. Consider the polynomial \(P(x)=x^2+5x-7\). You can take it as a fact that this polynomial has exactly two distinct roots. What is the sum of its roots? What about their product?
The polynomial \(P(x)=x^3+3x^2-7x+1\) has three distinct roots: \(a,b,\) and \(c\). What is the value of \(a^2+b^2+c^2\)?
So far we have discussed polynomials in one variable, i.e: with only an \(x\) as our variable. We can however, include as many as we want. For example, we can talk of a polynomial such as \[P(x,y)=x^2-y^2,\] where both \(x\) and \(y\) are variables. This is an example of an antisymmetric polynomial, which means that \(P(x,y)=-P(y,x)\) (i.e: switching \(x\) for \(y\) gives the original polynomial with a minus sign). Conversely, a polynomial \(Q(x,y)\) such that \(Q(x,y)=Q(y,x)\) is called symmetric. Show that every antisymmetric polynomial \(P(x,y)\) can be factored as \[P(x,y)=(x-y)Q(x,y),\] where \(Q(x,y)\) is a symmetric polynomial.
Find the remainder of dividing \(x^{100}-2x^{51}+1\) by \(x^2-1\). Try not to do a long calculation.