MathDB
Square prime

Source: Kyiv City MO 2024 Round 1, Problem 8.2

1/28/2024
Write the numbers from 11 to 1616 in the cells of a of a 4×44 \times 4 square so that: 1. Each cell contains exactly one number; 2. Each number is written exactly once; 3. For any two cells that are symmetrical with respect to any of the perpendicular bisectors of sides of the original 4×44 \times 4 square, the sum of numbers in them is a prime number
The figure below shows examples of such pairs of cells, sums of numbers in which have to be prime.
https://i.ibb.co/fqX05dY/Kyiv-MO-2024-Round-1-8-2.png
Proposed by Mykhailo Shtandenko
number theoryprime numbers
Congenial concyclicity

Source: Kyiv City MO 2024 Round 1, Problem 8.3

1/28/2024
The circle γ\gamma passing through the vertex AA of triangle ABCABC intersects its sides ABAB and ACAC for the second time at points XX and YY, respectively. Also, the circle γ\gamma intersects side BCBC at points DD and EE so that AD=AEAD = AE. Prove that the points B,X,Y,CB, X, Y, C lie on the same circle.
Proposed by Mykhailo Shtandenko
geometryConcyclic
Irreducible difference

Source: Kyiv City MO 2024 Round 1, Problem 9.1

1/28/2024
The difference of fractions 20242023−20232024\frac{2024}{2023} - \frac{2023}{2024} was represented as an irreducible fraction pq\frac{p}{q}. Find the value of pp.
number theoryalgebra
Answer to this NT might surprise you...

Source: Kyiv City MO 2024 Round 1, Problem 8.5

1/28/2024
Find the smallest positive integer nn that has at least 77 positive divisors 1=d1<d2<…<dk=n1 = d_1 < d_2 < \ldots < d_k = n, k≥7k \geq 7, and for which the following equalities hold:
d7=2d5+1 and d7=3d4−1d_7 = 2d_5 + 1\text{ and }d_7 = 3d_4 - 1
Proposed by Mykyta Kharin
number theoryDivisors
Maestro saves algebra

Source: Kyiv City MO 2024 Round 1, Problem 8.4

1/28/2024
Positive real numbers a1,a2,…,a2024a_1, a_2, \ldots, a_{2024} are arranged in a circle. It turned out that for any i=1,2,…,2024i = 1, 2, \ldots, 2024, the following condition holds: aiai+1<ai+2a_ia_{i+1} < a_{i+2}. (Here we assume that a2025=a1a_{2025} = a_1 and a2026=a2a_{2026} = a_2). What largest number of positive integers could there be among these numbers a1,a2,…,a2024a_1, a_2, \ldots, a_{2024}?
Proposed by Mykhailo Shtandenko
algebra
Is this algebra statement not nice?

Source: Kyiv City MO 2024 Round 1, Problem 10.5/11.4

1/28/2024
Find the smallest real number MM, for which {a}+{b}+{c}≤M\{a\}+\{b\}+\{c\}\leq M for any real positive numbers a,b,ca, b, c with abc=2024abc = 2024. Here {a}\{a\} denotes the fractional part of number aa.
Proposed by Fedir Yudin, Anton Trygub
algebra
Medium numbergame theory

Source: Kyiv City MO 2024 Round 1, Problem 9.3

1/28/2024
Petro and Vasyl play the following game. They take turns making moves and Petro goes first. In one turn, a player chooses one of the numbers from 11 to 20232023 that wasn't selected before and writes it on the board. The first player after whose turn the product of the numbers on the board will be divisible by 20232023 loses. Who wins if every player wants to win?
Proposed by Mykhailo Shtandenko
gamenumber theoryDivisibility
Strange divisibility

Source: Kyiv City MO 2024 Round 1, Problem 10.1

1/28/2024
Find all pairs of positive integers (a,b)(a, b) such that 4b−14b - 1 is divisible by 3a+13a + 1 and 3a−13a - 1 is divisible by 2b+12b + 1.
number theoryalgebraDivisibility
Smallest value of |253^m - 40^n|

Source: Kyiv City MO 2024 Round 1, Problem 9.5

1/28/2024
Find the smallest value of the expression ∣253m−40n∣|253^m - 40^n| over all pairs of positive integers (m,n)(m, n).
Proposed by Oleksii Masalitin
number theoryDiophantine equation
Hard numbergame theory

Source: Kyiv City MO 2024 Round 1, Problem 11.3

1/28/2024
Let n>1n>1 be a given positive integer. Petro and Vasyl play the following game. They take turns making moves and Petro goes first. In one turn, a player chooses one of the numbers from 11 to nn that wasn't selected before and writes it on the board. The first player after whose turn the product of the numbers on the board will be divisible by nn loses. Who wins if every player wants to win? Find answer for each n>1n>1.
Proposed by Mykhailo Shtandenko, Anton Trygub
gamenumber theorycombinatoricsDivisibility
Fantastic polynomial problem

Source: Kyiv City MO 2024 Round 1, Problem 10.4

1/28/2024
For a positive integer nn, does there exist a permutation of all its positive integer divisors (d1,d2,…,dk)(d_1 , d_2 , \ldots, d_k) such that the equation dkxk−1+…+d2x+d1=0d_kx^{k-1} + \ldots + d_2x + d_1 = 0 has a rational root, if:
a) n=2024n = 2024; b) n=2025n = 2025?
Proposed by Mykyta Kharin
algebrapolynomialnumber theory
Knights accuse each other in lying!

Source: Kyiv City MO 2024 Round 1, Problem 10.3

1/28/2024
There are 20252025 people living on the island, each of whom is either a knight, i.e. always tells the truth, or a liar, which means they always lie. Some of the inhabitants of the island know each other, and everyone has at least one acquaintance, but no more than three. Each inhabitant of the island claims that there are exactly two liars among his acquaintances.
a) What is the smallest possible number of knights among the inhabitants of the island? b) What is the largest possible number of knights among the inhabitants of the island?
Proposed by Oleksii Masalitin
combinatoricslying
It&#039;s a trapezoid

Source: Kyiv City MO 2024 Round 1, Problem 11.2

1/28/2024
ABCDABCD is a trapezoid with BC∥ADBC\parallel AD and BC=2ADBC = 2AD. Point MM is chosen on the side CDCD such that AB=AMAB = AM. Prove that BM⊥CDBM \perp CD.
Proposed by Bogdan Rublov
geometrytrapezoid
Funky number theory

Source: Kyiv City MO 2024 Round 1, Problem 11.5

1/28/2024
Find all functions f:N→Nf : \mathbb{N} \to \mathbb{N} such that for any positive integers m,nm, n the number (f(m))2+2mf(n)+f(n2)(f(m))^2+ 2mf(n) + f(n^2) is the square of an integer.
Proposed by Fedir Yudin
functional equationnumber theoryFunctional Equations
Nondivisible power sums

Source: Kyiv City MO 2024 Round 2, Problem 9.2

2/4/2024
You are given a positive integer n>1n > 1. What is the largest possible number of integers that can be chosen from the set {1,2,3,…,2n}\{1, 2, 3, \ldots, 2^n\} so that for any two different chosen integers a,ba, b, the value ak+bka^k + b^k is not divisible by 2n2^n for any positive integer kk?
Proposed by Oleksii Masalitin
number theory
At least one nonnegative

Source: Kyiv City MO 2024 Round 2, Problem 7.1

2/4/2024
Prove that for any real numbers x,y,zx, y, z at least one of numbers x2+y+14,y2+z+14,z2+x+14x^2 + y + \frac{1}{4}, y^2 + z + \frac{1}{4}, z^2 + x + \frac{1}{4} is nonnegative.
Proposed by Oleksii Masalitin
algebrainequalities
Twice meaner

Source: Kyiv City MO 2024 Round 1, Problem 11.1

1/28/2024
Four positive integers a,b,c,da, b, c, d satisfy the condition: a<b<c<da < b < c < d. For what smallest possible value of dd could the following condition be true: the arithmetic mean of numbers a,b,ca, b, c is twice smaller than the arithmetic mean of numbers a,b,c,da, b, c, d?
algebramean
Anti-GCD problem

Source: Kyiv City MO 2024 Round 2, Problem 7.2/8.1

2/4/2024
You are given a positive integer nn. What is the largest possible number of numbers that can be chosen from the set {1,2,…,2n}\{1, 2, \ldots, 2n\} so that there are no two chosen numbers x>yx > y for which x−y=(x,y)x - y = (x, y)?
Here (x,y)(x, y) denotes the greatest common divisor of x,yx, y.
Proposed by Anton Trygub
GCDnumber theory
Dividing circle into parts with equal sums

Source: Kyiv City MO 2024 Round 2, Problem 7.3

2/4/2024
20242024 ones and 20242024 twos are arranged in a circle in some order. Is it always possible to divide the circle into
a) two (contiguous) parts with equal sums? b) three (contiguous) parts with equal sums?
Proposed by Fedir Yudin
combinatoricsconstruction
8-graders solving algebra from Shortlist (unintended)

Source: Kyiv City MO 2024 Round 2, Problem 8.2

2/4/2024
Find the smallest positive integer nn for which one can select nn distinct real numbers such that each of them is equal to the sum of some two other selected numbers.
Proposed by Anton Trygub
algebraconstruction
(Angle) Chasing good geometry

Source: Kyiv City MO 2024 Round 2, Problem 8.3

2/4/2024
Let ω\omega denote the circumscribed circle of an acute-angled △ABC\triangle ABC with AB≠BCAB \neq BC. Let A′A' be the point symmetric to the point AA with respect to the line BCBC. The lines AA′AA' and A′CA'C intersect ω\omega for the second time at points DD and EE, respectively. Let the lines AEAE and BDBD intersect at point PP. Prove that the line A′PA'P is tangent to the circumscribed circle of △A′BC\triangle A'BC.
Proposed by Oleksii Masalitin
geometrycircumcircle
Genius strikes again: nobody solved this in contest!

Source: Kyiv City MO 2024 Round 2, Problem 7.4

2/4/2024
Points XX and YY are chosen inside an acute-angled triangle ABCABC with altitude ADAD so that ∠BXA+∠ACB=180∘,∠CYA+∠ABC=180∘\angle BXA + \angle ACB = 180^\circ , \angle CYA + \angle ABC = 180^\circ, and CD+AY=BD+AXCD + AY = BD + AX. Point MM is chosen on the ray BXBX so that XX lies on segment BMBM and XM=ACXM = AC, and point NN is chosen on the ray CYCY so that YY lies on segment CNCN and YN=ABYN = AB. Prove that AM=ANAM = AN.
Proposed by Mykhailo Shtandenko
geometry
Economic crisis :(

Source: Kyiv City MO 2024 Round 2, Problem 8.4/9.3

2/4/2024
In a certain magical country, there are banknotes in denominations of 20,21,22,…2^0, 2^1, 2^2, \ldots UAH. Businessman Victor has to make cash payments to 4444 different companies totaling 4400044000 UAH, but he does not remember how much he has to pay to each company. What is the smallest number of banknotes Victor should withdraw from an ATM (totaling exactly 4400044000 UAH) to guarantee that he would be able to pay all the companies without leaving any change?
Proposed by Oleksii Masalitin
combinatoricsnumber theory
System of equations: easy

Source: Kyiv City MO 2024 Round 2, Problem 9.1

2/4/2024
Solve the following system of equations in real numbers: {x2=y2+z2,x2023=y2023+z2023,x2025=y2025+z2025.\left\{\begin{array}{l}x^2=y^2+z^2,\\x^{2023}=y^{2023}+z^{2023},\\x^{2025}=y^{2025}+z^{2025}.\end{array}\right. Proposed by Mykhailo Shtandenko, Anton Trygub
algebrasystem of equations
Beautiful geometry from Kyiv MO

Source: Kyiv City MO 2024 Round 2, Problem 9.4

2/4/2024
Let BDBD be an altitude of △ABC\triangle ABC with AB<BCAB < BC and ∠B>90∘\angle B > 90^\circ. Let MM be the midpoint of ACAC, and point KK be symmetric to point DD with respect to point MM. A perpendicular drawn from point MM to the line BCBC intersects line ABAB at point LL. Prove that ∠MBL=∠MKL\angle MBL = \angle MKL.
Proposed by Oleksandra Yakovenko
geometry