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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 \(k\)?

Author: A.K. Tolpygo

An irrational number \(\alpha\), where \(0 <\alpha <\frac 12\), is given. It defines a new number \(\alpha_1\) as the smaller of the two numbers \(2\alpha\) and \(1 - 2\alpha\). For this number, \(\alpha_2\) is determined similarly, and so on.

a) Prove that for some \(n\) the inequality \(\alpha_n <3/16\) holds.

b) Can it be that \(\alpha_n> 7/40\) for all positive integers \(n\)?

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 \((x - x_0)^2\).

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?

When Robinson Crusoe and his friend Friday learned about divisibility rules, Friday decided to proposed his own rule:

if a number is divisible by \(27\), then the sum of its digits is also divisible by \(27\).

Was Friday right?

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.