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Seven robbers are dividing a bag of coins of various denominations. It turned out that the sum could not be divided equally between them, but if any coin is set aside, the rest could be divided so that every robber would get an equal part. Prove that the bag cannot contain \(100\) coins.

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).

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.

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?