Solving the problem: “What is the solution of the expression \(x^{2000} + x^{1999} + x^{1998} + 1000x^{1000} + 1000x^{999} + 1000x^{998} + 2000x^3 + 2000x^2 + 2000x + 3000\) (\(x\) is a real number) if \(x^2 + x + 1 = 0\)?”, Vasya got the answer of 3000. Is Vasya right?
Prove that amongst the numbers of the form \[19991999\dots 19990\dots 0\] – that is 1999 a number of times, followed by a number of 0s – there will be at least one divisible by 2001.
Let \(M\) be the point of intersection of the medians of the triangle \(ABC\), and \(O\) an arbitrary point on a plane. Prove that \[OM^2 = 1/3 (OA^2 + OB^2 + OC^2) - 1/9 (AB^2 + BC^2 + AC^2).\]
Three non-coplanar vectors are given. Is it possible to find a fourth vector perpendicular to the three vectors given?
Find the volume of an inclined triangular prism whose base is an equilateral triangle with sides equal to a if the side edge of the prism is equal to the side of the base and is inclined to the plane of the base at an angle of \(60^{\circ}\).
Is it possible to cut out such a hole in a sheet of paper through which a person could climb through?
a) Prove that within any 6 whole numbers there will be two that have a difference between them that is a multiple of 5.
b) Will this statement remain true if instead of the difference we considered the total?
Is the number \(10^{2002} + 8\) divisible by 9?
Is the sum of the numbers \(1 + 2 + 3 + \dots + 1999\) divisible by 1999?
In an ordinary set of dominoes, there are 28 tiles. How many tiles would a set of dominoes contain if the values indicated on the tiles did not range from 0 to 6, but from 0 to 12?