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 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 \(3\).
Find the smallest \(k\) such that \(k!\) (\(k!= k\times(k-1)\times \ldots \times 1\)) is divisible by \(2024\).
While studying numbers and its properites, Robinson came across a 3-digit prime number with the last digit being equal to the sum of the first two digits. What was the last digit of that number if among the number did not have any zeros among it’s digits?
When Robinson Crusoe’s friend and assistant named Friday learned about divisibility rules, he was so impressed that he proposed his own rule:
a number is divisible by 27 if the sum of it’s digits is divisible by 27.
Was he right?