Is there a sequence of natural numbers in which every natural number occurs exactly once, and for any \(k = 1, 2, 3, \dots\) the sum of the first \(k\) terms of the sequence is divisible by \(k\)?
Prove that there are no natural numbers \(a\) and \(b\) such that \(a^2 - 3b^2 = 8\).
Prove that if \((p, q) = 1\) and \(p/q\) is a rational root of the polynomial \(P (x) = a_nx^n + \dots + a_1x + a_0\) with integer coefficients, then
a) \(a_0\) is divisible by \(p\);
b) \(a_n\) is divisible by \(q\).
What has a greater value: \(300!\) or \(100^{300}\)?
Prove that if \(x_0^4 + a_1x_0^3 + a_2x_0^2 + a_3x_0 + a_4\) and \(4x_0^3 + 3a_1x_0^2 + 2a_2x_0 + a_3 = 0\) then \(x^4 + a_1x^3 + a_2x^2 + a_3x + a_4\) is divisible by \((x - x_0)^2\).
Does there exist a number \(h\) such that for any natural number \(n\) the number \(\lfloor h \times 2021^n\rfloor\) is not divisible by \(\lfloor h \times 2021^{n-1}\rfloor\)?
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 \(3\): the number is divisible by \(3\) if and only if the sum of its digits is divisible by \(3\).
Sophia is playing the following game: she chooses a whole number, and then she writes down the product of all the numbers from \(1\) up to the number she chose. For example, if she chooses \(5\), then she writes down \(1\times 2 \times 3 \times 4 \times 5\). What is the smallest number she can choose for her game, such that the result she gets in the end is divisible by \(2024\)?
While studying numbers and their properties, Robinson came across a three-digit prime number whose last digit equals the sum of the first two digits. What are the options for the last digit of this number, given that none of its digits is zero?