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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).

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 \(O\)).

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?