Blondes, brunettes, redheads and brown-haired women participate in the Olympiad. There are twice as many redheads as brown-haired. Blondes and redheads make up a quarter of the total number of participants, and Brown-haired and Blondes one fifth part. Prove that the number of Brunettes is divisible by $7$.
2022 Assara - South Russian Girl's MO
Juniors
day 1
There are $2022$ natural numbers written in a row. Product of any two adjacent numbers is a perfect cube. Prove that the product of the two extremes is also a perfect cube.
For what $n$ can the vertices of a regular $n$-gon be connected in a $n$-link closed polyline so that such a polyline does not have three equal links?
Nadya has $2022$ cards, each with a number one or seven written on it. It is known that there are both cards.Nadya looked at all possible $2022$-digit numbers that can be composed from all these cards. What is the largest value that can take the greatest common divisor of all these numbers?
day 2
Find all pairs of prime numbers $p, q$ such that the number $pq + p - 6$ is also prime.
The cells of the $9 \times 9$ table are colored black and white. It turned out, that there were $k$ rows, in each of which there are more black cells than white ones white, and there were $k$ columns, each of which contained more than black. At what highest $ k$ is this possible?
Find out which of the two numbers is greater: $$\dfrac{2}{2 +\dfrac{2}{2 +\dfrac{2}{... +\dfrac{2}{2+\frac22}}}} \,\,\, \text{or} \,\,\, \dfrac{3}{3 +\dfrac{3}{3 +\dfrac{3}{... +\dfrac{3}{3+\frac33}}}}$$(Each expression has $2022$ fraction signs.)
About the convex hexagon $ABCDEF$ it is known that $AB = BC =CD = DE = EF = FA$ and $AD = BE = CF$. Prove that the diagonals $AD$, $BE$, $CF$ intersect at one point.
Seniors
day 1
Given three natural numbers $a$, $b$ and $c$. It turned out that they are coprime together. And their least common multiple and their product are perfect squares. Prove that $a$, $b$ and $c$ are perfect squares.
Numbers $1, 2, 3, . . . , 100$ are arranged in a circle in some order. A good pair is a pair of numbers of the same parity, between which there are exactly $3$ numbers. What is the smallest possible number good pairs?
In a convex quadrilateral $ABCD$, angles $B$ and $D$ are right angles. On on sides $AB$, $BC$, $CD$, $DA$ points $K$, $L$, $M$, $N$ are taken respectively so that $KN \perp AC$ and $LM \perp AC$. Prove that $KM$, $LN$ and $AC$ intersect at one point.
Alina knows how to twist a periodic decimal fraction in the following way: she finds the minimum preperiod of the fraction, then takes the number that makes up the period and rearranges the last one in it digit to the beginning of the number. For example, from the fraction, $0.123(56708)$ she will get $0.123(85670)$. What fraction will Alina get from fraction $\frac{503}{2022}$ ?
day 2
same sa Juniors p7 - 5
There are $2022$ numbers arranged in a circle $a_1, a_2, . . ,a_{2022}$. It turned out that for any three consecutive $a_i$, $a_{i+1}$, $a_{i+2}$ the equality $a_i =\sqrt2 a_{i+2} - \sqrt3 a_{i+1}$. Prove that $\sum^{2022}_{i=1} a_ia_{i+2} = 0$, if we know that $a_{2023} = a_1$, $a_{2024} = a_2$.
In a $7\times 7\times 7$ cube, the unit cubes are colored white, black and gray colors so that for any two colors the number of cubes of these two colors are different. In this case, $N$ parallel rows of $7$ cubes were found, each of which there are more white cubes than gray and than black. Likewise, there were $N$ parallel rows of $7$ cubes, each of which contained gray there are more cubes than white and than black, and there are also N parallel rows of $7$ cubes, each of which contains more black cubes than white ones and than gray ones. What is the largest $N$ for which this is possible?
In parallelogram $ABCD$, angle $A$ is acute. Let $X$ be a point, symmetrical to point $C$ wrt to straight line $AD$, $Y$ is a point symmetrical to the point $C$ wrt point $D$, and $M$ is the intersection point of $AC$ and $BD$. It turned out, that the circumcircles of triangles $BMC$ and $AXY$ are tangent internally. Prove that $AM = AB$.