Prove that for any odd natural number, \(a\), there exists a natural number, \(b\), such that \(2^b - 1\) is divisible by \(a\).
Prove that for any positive integer \(n\), it is always possible to find a number, consisting of the digits \(1\) and \(2,\) that is divisible by \(2^n\). (For example, \(2\) is divisible by \(2\), \(12\) is divisible by \(4,\) \(112\) is divisible by \(8,\) \(2112\) is divisible by \(16\) and so on...).
A sequence of natural numbers \(a_1 < a_2 < a_3 < \dots < a_n < \dots\) is such that each natural number is either a term in the sequence, can be expressed as the sum of two terms in the sequence, or perhaps the same term twice. Prove that \(a_n \leq n^2\) for any \(n=1, 2, 3,\dots\)
Out of the given numbers 1, 2, 3, ..., 1000, find the largest number \(m\) that has this property: no matter which \(m\) of these numbers you delete, among the remaining \(1000 - m\) numbers there are two, of which one is divisible by the other.
Let’s call a natural number good if in its decimal record we have the numbers 1, 9, 7, 3 in succession, and bad if otherwise. (For example, the number 197,639,917 is bad and the number 116,519,732 is good.) Prove that there exists a positive integer \(n\) such that among all \(n\)-digit numbers (from \(10^{n-1}\) to \(10^{n-1}\)) there are more good than bad numbers.
Try to find the smallest possible \(n\).
An infinite sequence of digits is given. Prove that for any natural number \(n\) that is relatively prime with a number 10, you can choose a group of consecutive digits, which when written as a sequence of digits, gives a resulting number written by these digits which is divisible by \(n\).
What has a greater value: \(300!\) or \(100^{300}\)?
Prove that, for any integer \(n\), among the numbers \(n, n + 1, n + 2, \dots , n + 9\) there is at least one number that is mutually prime with the other nine numbers.
There are 13 weights, each weighing an integer number of grams. It is known that any 12 of them can be divided into two cups of weights, six weights on each one, which will come to equilibrium. Prove that all the weights have the same weight.
If we are given any 100 whole numbers then amongst them it is always possible to choose one, or several of them, so that their sum gives a number divisible by 100. Prove that this is the case.