Problems

In the diagram below, there are nine discs - each is black on one side, and white on the other side. Two have black face-up right now. Your task is to remove all of the discs by making a series of the following moves. Each move includes choosing a black disc, flipping over its neighbours${}^{\ast }$ and removing that black disc. Discs are ‘neighbours’ if they’re adjacent at the beginning - removing a disc creates a gap, so that at later stages, a disc may have two, one or even zero neighbours left. $$\circ \circ \circ \bullet \circ \circ \circ \circ \bullet$$ Show that this task is impossible.

Draw the plane tiling with regular hexagons.

Let $\sigma (n)$ be the sum of the divisors of $n$. For example, $\sigma (12)=1+2+3+4+6+12=28$. We use $\gamma$ to denote the Euler-Mascheroni constant - one way to define this is as $\gamma :=\underset{n\to \mathrm{\infty }}{lim}(\sum _{k=1}^n\frac{1}{n}-\log n)$.

Prove that $\sigma (n)<e^{\gamma }n\log \log n$ for all integers $n>5040$.

Let $a$ and $b$ be two different $9$-digit numbers. It is known that each one of them contains all of the digits $1,2,...9$. Find the maximal value of $gcd(a,b)$.

Take a regular dodecahedron as in the image. It has $12$ regular pentagons as its faces, $30$ edges, and $20$ vertices. We can cut it with planes in various ways and the cut will be a polygon on a plane. Find out how many ways there are to cut a dodecahedron with a plane so that the polygon obtained is a regular hexagon.

For an odd number $N$ denote by $A$ the minimal positive difference between prime divisors of $N$, denote by $B$ the minimal positive difference between composite divisors of $N$. Usually we have $A<B$, but can we have $A>B$? (Disregard numbers such as $15$ where one of $A$ or $B$ is not defined)

Let $n$ be an integer bigger than $1$, and $p$ a prime number. Suppose that $n$ divides $p-1$ and $p$ divides $n^3-1$. Prove that $4p-3$ is a square number.

Let $n$ be a composite number. Arrange the factors of $n$ greater than $1$ in a circle. When can this be done such that neighbours in the circle are never coprime?

Let $x$, $y$, $z$ and $w$ be non-negative integers. Find all solutions to $2^x{3}^y-5^z{7}^w=1$.

A natural number $N$ is called perfect if it equals the sum of its divisors, except for $N$ itself. Prove that if $2^r-1$ is prime, then $(2^r-1)2^{r-1}$ is a perfect number. Are there any odd perfect numbers?