There are \(n\) integers. Prove that among them either there are several numbers whose sum is divisible by \(n\) or there is one number divisible by \(n\) itself.
Convert 2000 seconds into minutes and seconds.
Prove that any \(n\) numbers \(x_1,\dots , x_n\) that are not pairwise congruent modulo \(n\), represent a complete system of residues, modulo \(n\).
Prove that for any natural number there is a multiple of it, the decimal notation of which consists of only 0 and 1.
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.
All integers from 1 to \(2n\) are written in a row. Then, to each number, the number of its place in the row is added, that is, to the first number 1 is added, to the second – 2, and so on.
Prove that among the sums obtained there are at least two that give the same remainder when divided by \(2n\).
The sum of 100 natural numbers, each of which is no greater than 100, is equal to 200. Prove that it is possible to pick some of these numbers so that their sum is equal to 100.
Prove that in any group of 7 natural numbers – not necessarily consecutive – it is possible to choose three numbers such that their sum is divisible by 3.
a) We are given two cogs, each with 14 teeth. They are placed on top of one another, so that their teeth are in line with one another and their projection looks like a single cog. After this 4 teeth are removed from each cog, the same 4 teeth on each one. Is it always then possible to rotate one of the cogs with respect to the other so that the projection of the two partially toothless cogs appears as a single complete cog? The cogs can be rotated in the same plane, but cannot be flipped over.
b) The same question, but this time two cogs of 13 teeth each from which 4 are again removed?
Is it possible to arrange the numbers 1, 2, ..., 60 in a circle in such an order that the sum of every two numbers, between which lies one number, is divisible by 2, the sum of every two numbers between which lie two numbers, is divisible by 3, the sum of every two numbers between which lie six numbers, is divisible by 7?