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?
Prove that: \[a_1 a_2 a_3 \cdots a_{n-1}a_n \times 10^3 \equiv a_{n-1} a_n \times 10^3 \pmod4,\] where \(n\) is a natural number and \(a_i\) for \(i=1,2,\ldots, n\) are the digits of some number.
How many different four-digit numbers, divisible by 4, can be made up of the digits 1, 2, 3 and 4,
a) if each number can occur only once?
b) if each number can occur several times?
Is the number 12345678926 square?
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.
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 \(125\)?
Is \(100! = 100\times 99 \times ...\times 2\) divisible by \(2^{100}\)?