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.
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?
You are given 12 different whole numbers. Prove that it is possible to choose two of these whose difference is divisible by 11.