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.
If we are given any 100 whole numbers then amongst them it is always possible to choose one, or several of them, so that their sum gives a number divisible by 100. Prove that this is the case.
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).
Find the locus of points whose coordinates \((x, y)\) satisfy the relation \(\sin(x + y) = 0\).
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 \(x_1 \geq x_3 \geq x_2\).
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 \(O\)).
The expression \(ax^2+bx+c\) is an exact fourth power for all integers \(x\). Prove that \(a=b=0\).
The numbers \(\lfloor a\rfloor, \lfloor 2a\rfloor, \dots , \lfloor Na\rfloor\) are all different, and the numbers \(\lfloor 1/a\rfloor, \lfloor 2/a\rfloor,\dots , \lfloor M/a\rfloor\) are also all different. Find all such \(a\).
2022 points are selected from a cube, whose edge is equal to 13 units. Is it possible to place a cube with edge of 1 unit in this cube so that there is not one selected point inside it?
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.