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?
What figure should I put in place of the “?” in the number \(888 \dots 88\,?\,99 \dots 999\) (eights and nines are written 50 times each) so that it is divisible by 7?
Find the number of solutions in natural numbers of the equation \(\lfloor x / 10\rfloor = \lfloor x / 11\rfloor + 1\).
A numerical sequence is defined by the following conditions: \[a_1 = 1, \quad a_{n+1} = a_n + \lfloor \sqrt{a_n}\rfloor .\]
Prove that among the terms of this sequence there are an infinite number of complete squares.
Prove the divisibility rule for \(25\): a number is divisible by \(25\) if and only if the number made by the last two digits of the original number is divisible by \(25\);
Can you come up with a divisibility rule for \(125\)?
Which of the following numbers are divisible by \(11\) and which are not? \[121,\, 143,\, 286, 235, \, 473,\, 798, \, 693,\, 576, \,748\] Can you write down and prove a divisibility rule which helps to determine if a three digit number is divisible by \(11\)?
Is \(100! = 100\times 99 \times ...\times 2\) divisible by \(2^{100}\)?
Does there exist a power of \(3\) that ends in \(0001\)?