Problems

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Found: 12

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.

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?

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.