Solve the inequality: \(|x + 2000| <|x - 2001|\).
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.
Is it possible to cut out such a hole in a sheet of paper through which a person could climb through?
The best student in the class, Katie, made up a huge number, writing out in a row all of the natural numbers from 1 to 500: \[123 \dots 10111213 \dots 499500.\] The second-best student, Tom, erased the first 500 digits of this number. What do you think, what number does the remaining number begin with?
Fred and George together with their mother were decorating the Christmas tree. So that they would not fight, their mother gave each brother the same number of decorations and branches. Fred tried to hang one decoration on each branch, but he needed one more branch for his last decoration. George tried to hang two toys on each branch, but one branch was empty. What do you think, how many branches and how many decorations did the mother give to her sons?
The farmer must transport across a river a wolf, a goat and a cabbage. The boat accommodates one person, and with him/her either a wolf, a goat, or a cabbage. If you leave the goat and the wolf unattended, the wolf will eat the goat. If you leave cabbage and goat without supervision, the goat will eat the cabbage. How can the farmer transport his cargo across the river?
There is a 12-litre barrel filled with beer, and two empty kegs of 5 and 8 litres. Try using these kegs to:
a) divide the beer into two parts of 3 and 9 litres;
b) divide the beer into two equal parts.
There are scales without weights and 3 identical in appearance coins, one of which is fake: it is lighter than the real ones (the real coins are of the same weight). How many weightings are needed to determine the counterfeit coin? Solve the same problem in the cases where there are 4 coins and 9 coins.
We have scales without weights and 3 identical in appearance coins. One of the coins is fake, and it is not known whether it is lighter or heavier than the real coins (note that all real coins are of the same weight). How many weighings are needed to determine the counterfeit coin? Solve the same problem in the cases where there are 4 coins and 9 coins.