All of the integers from 1 to 64 are written in an \(8 \times 8\) table. Prove that in this case there are two adjacent numbers, the difference between which is not less than 5. (Numbers that are in cells which share a common side are called adjacent).
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 2001 whole numbers there will be two whose difference is divisible by 2000.
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.
Find all of the natural numbers that, when divided by 7, have the same remainder and quotient.
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?
Will the quotient or the remainder change if a divided number and the divisor are increased by 3 times?
A group of numbers \(A_1, A_2, \dots , A_{100}\) is created by somehow re-arranging the numbers \(1, 2, \dots , 100\).
100 numbers are created as follows: \[B_1=A_1,\ B_2=A_1+A_2,\ B_3=A_1+A_2+A_3,\ \dots ,\ B_{100} = A_1+A_2+A_3\dots +A_{100}.\]
Prove that there will always be at least 11 different remainders when dividing the numbers \(B_1, B_2, \dots , B_{100}\) by 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.