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 for a real positive \(\alpha\) and a positive integer \(d\), \(\lfloor \alpha / d\rfloor = \lfloor \lfloor \alpha\rfloor / d\rfloor\) is always satisfied.
Prove that if \(p\) is a prime number and \(1 \leq k \leq p - 1\), then \(\binom{p}{k}\) is divisible by \(p\).
Prove that if \(p\) is a prime number, then \((a + b)^p - a^p - b^p\) is divisible by \(p\) for any integers \(a\) and \(b\).
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 \(\lfloor h \times 2021^{n-1}\rfloor\)?
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?
One day Friday multiplied all the numbers from 1 to 100. The product appeared to be a pretty large number, and he added all the digits of that number to receive a new smaller number. Even then he did not think the number was small enough, and added all the digits again to receive a new number. He continued this process of adding all the digits of the newly obtained number again and again, until finally he received a one-digit number. Can you tell what number was it?
Robinson Crusoe’s friend Friday was looking at \(3\)-digit numbers with the same first and third digits. He soon noticed that such number is divisible by \(7\) if the sum of the second and the third digits is divisible by \(7\). Prove that he was right.
2016 digits are written in a circle. It is known, that if you make a number reading the digits clockwise, starting from some particular place, then the resulting 2016-digit number is divisible by 27. Show that if you start from some other place, and moving clockwise make up another 2016-digit number, then this new number is also divisible by 27.