Show that any natural number has the same remainder when divided by \(3\) as the sum of its digits.
Show that \(n^3 + 2n\) is divisible by \(3\) for a natural \(n\).
Prove that if \(a^3- b^3\), for \(a\) and \(b\) natural, is divisible by \(3\), then it is divisible by \(9\).
Does there exist a natural number which, when divided by the sum of its digits, gives a quotient and remainder both equal to the number 2011?
It is known that \[35! = 10333147966386144929 * 66651337523200000000.\] Find the number replaced by an asterisk.
Without calculating the answer to \(2^{30}\), prove that it contains at least two identical digits.