There are 13 weights, each weighing an integer number of grams. It is known that any 12 of them can be divided into two cups of weights, six weights on each one, which will come to equilibrium. Prove that all the weights have the same weight.
Numbers \(1, 2, 3, \dots , 101\) are written out in a row in some order. Prove that one can cross out 90 of them so that the remaining 11 will be arranged in their magnitude (either increasing or decreasing).
The equations \[ax^2 + bx + c = 0 \tag{1}\] and \[- ax^2 + bx + c \tag{2}\] are given. Prove that if \(x_1\) and \(x_2\) are, respectively, any roots of the equations (1) and (2), then there is a root \(x_3\) of the equation \(\frac 12 ax^2 + bx + c\) such that either \(x_1 \leq x_3 \leq x_2\) or \(x_1 \geq x_3 \geq x_2\).
Prove that if \(x_0^4 + a_1x_0^3 + a_2x_0^2 + a_3x_0 + a_4\) and \(4x_0^3 + 3a_1x_0^2 + 2a_2x_0 + a_3 = 0\) then \(x^4 + a_1x^3 + a_2x^2 + a_3x + a_4\) is divisible by \((x - x_0)^2\).
The segment \(OA\) is given. From the end of the segment \(A\) there are 5 segments \(AB_1, AB_2, AB_3, AB_4, AB_5\). From each point \(B_i\) there can be five more new segments or not a single new segment, etc. Can the number of free ends of the constructed segments be 1001? By the free end of a segment we mean a point belonging to only one segment (except point \(O\)).
Solve the equation \(x^3 - \lfloor x\rfloor = 3\).
There is a system of equations \[\begin{aligned} * x + * y + * z &= 0,\\ * x + * y + * z &= 0,\\ * x + * y + * z &= 0. \end{aligned}\] Two people alternately enter a number instead of a star. Prove that the player that goes first can always ensure that the system has a non-zero solution.
There are two sets of numbers made up of 1s and \(-1\)s, and in each there are 2022 numbers. Prove that in some number of steps it is possible to turn the first set into the second one if for each step you are allowed to simultaneously change the sign of any 11 numbers of the starting set. (Two sets are considered the same if they have the same numbers in the same places.)
Given \(n\) points that are connected by segments so that each point is connected to some other and there are no two points that would be connected in two different ways. Prove that the total number of segments is \(n - 1\).
A system of points connected by segments is called “connected” if from each point one can go to any other one along these segments. Is it possible to connect five points to a connected system so that when erasing any segment, exactly two connected points systems are formed that are not related to each other? (We assume that in the intersection of the segments, the transition from one of them to another is impossible).