The triangle \(C_1C_2O\) is given. Within it the bisector \(C_2C_3\) is drawn, then in the triangle \(C_2C_3O\) – bisector \(C_3C_4\) and so on. Prove that the sequence of angles \(\gamma_n = C_{n + 1}C_nO\) tends to a limit, and find this limit if \(C_1OC_2 = \alpha\).
Prove that, for any integer \(n\), among the numbers \(n, n + 1, n + 2, \dots , n + 9\) there is at least one number that is mutually prime with the other nine numbers.
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).
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?
30 pupils in years 7 to 11 took part in the creation of 40 maths problems. Every possible pair of pupils in the same year created the same number of problems. Every possible pair of pupils in different years created a different number of problems. How many pupils created exactly one problem?
The number \(A\) is divisible by \(1, 2, 3, \dots , 9\). Prove that if \(2A\) is presented in the form of a sum of some natural numbers smaller than 10, \(2A= a_1 +a_2 +\dots +a_k\), then we can always choose some of the numbers \(a_1, a_2, \dots , a_k\) so that the sum of the chosen numbers is equal to \(A\).