MathDB
Problems
Contests
National and Regional Contests
Ukraine Contests
Official Ukraine Selection Cycle
Kyiv City MO
2021 Kyiv City MO
2021 Kyiv City MO Round 1
2021 Kyiv City MO Round 1
Part of
2021 Kyiv City MO
Subcontests
(23)
11.5
1
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Sum of powers
For positive integers
m
,
n
m, n
m
,
n
define the function
f
n
(
m
)
=
1
2
n
+
2
2
n
+
3
2
n
+
…
+
m
2
n
f_n(m) = 1^{2n} + 2^{2n} + 3^{2n} + \ldots +m^{2n}
f
n
(
m
)
=
1
2
n
+
2
2
n
+
3
2
n
+
…
+
m
2
n
. Prove that there are only finitely many pairs of positive integers
(
a
,
b
)
(a, b)
(
a
,
b
)
such that
f
n
(
a
)
+
f
n
(
b
)
f_n(a) + f_n(b)
f
n
(
a
)
+
f
n
(
b
)
is a prime number.Proposed by Nazar Serdyuk
11.4
1
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Another amazing inequality
For positive real numbers
a
,
b
,
c
a, b, c
a
,
b
,
c
with sum
3
2
\frac{3}{2}
2
3
, find the smallest possible value of the following expression:
a
3
b
c
+
b
3
c
a
+
c
3
a
b
+
1
a
b
c
\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} + \frac{1}{abc}
b
c
a
3
+
c
a
b
3
+
ab
c
3
+
ab
c
1
Proposed by Serhii Torba
11.2
1
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Skew knights
Chess piece called skew knight, if placed on the black square, attacks all the gray squares. https://i.ibb.co/HdTDNjN/Kyiv-MO-2021-Round-1-11-2.pngWhat is the largest number of such knights that can be placed on the
8
×
8
8\times 8
8
×
8
chessboard without them attacking each other?Proposed by Arsenii Nikolaiev
11.1
1
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Cossacks gossip
N
N
N
cossacks split into
3
3
3
groups to discuss various issues with their friends. Cossack Taras moved from the first group to the second, cossack Andriy moved from the second to the third, and cossack Ostap - from the third group to the first. It turned out that the average height of the cossacks in the first group decreased by
8
8
8
cm, while in the second and third groups it increased by
5
5
5
cm and
8
8
8
cm, respectively. What is
N
N
N
, if it is known that there were
9
9
9
cossacks in the first group?
10.5
1
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Largest power of $3$ dividing a_101
The sequence
(
a
n
)
(a_n)
(
a
n
)
is such that
a
n
+
1
=
(
a
n
)
n
+
n
+
1
a_{n+1} = (a_n)^n + n + 1
a
n
+
1
=
(
a
n
)
n
+
n
+
1
for all positive integers
n
n
n
, where
a
1
a_1
a
1
is some positive integer. Let
k
k
k
be the greatest power of
3
3
3
by which
a
101
a_{101}
a
101
is divisible. Find all possible values of
k
k
k
.Proposed by Kyrylo Holodnov
10.4
1
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Beautiful inequality
Positive real numbers
a
,
b
,
c
a, b, c
a
,
b
,
c
satisfy
a
2
+
b
2
+
c
2
+
a
+
b
+
c
=
6
a^2 + b^2 + c^2 + a + b + c = 6
a
2
+
b
2
+
c
2
+
a
+
b
+
c
=
6
. Prove the following inequality:
2
(
1
a
2
+
1
b
2
+
1
c
2
)
≥
3
+
1
a
+
1
b
+
1
c
2(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}) \geq 3 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
2
(
a
2
1
+
b
2
1
+
c
2
1
)
≥
3
+
a
1
+
b
1
+
c
1
Proposed by Oleksii Masalitin
10.3
1
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Circles and parallelogram
Circles
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
with centers at points
O
1
O_1
O
1
and
O
2
O_2
O
2
intersect at points
A
A
A
and
B
B
B
. Let point
C
C
C
be such that
A
O
2
C
O
1
AO_2CO_1
A
O
2
C
O
1
is a parallelogram. An arbitrary line is drawn through point
A
A
A
, which intersects the circles
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
at points
X
X
X
and
Y
Y
Y
, respectively. Prove that
C
X
=
C
Y
CX = CY
CX
=
C
Y
.Proposed by Oleksii Masalitin
10.2
1
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Almost magic square with a twist
The
1
×
1
1 \times 1
1
×
1
cells located around the perimeter of a
4
×
4
4 \times 4
4
×
4
square are filled with the numbers
1
,
2
,
…
,
12
1, 2, \ldots, 12
1
,
2
,
…
,
12
so that the sums along each of the four sides are equal. In the upper left corner cell is the number
1
1
1
, in the upper right - the number
5
5
5
, and in the lower right - the number
11
11
11
. https://i.ibb.co/PM0ry1D/Kyiv-City-MO-2021-Round-1-10-2.pngUnder these conditions, what number can be located in the last corner cell?Proposed by Mariia Rozhkova
10.1
1
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Hardest problem on Kyiv City MO 2021. Possibly ever
Prove the following inequality:
sin
1
+
sin
3
+
…
+
sin
2021
>
2
sin
1011
2
3
\sin{1} + \sin{3} + \ldots + \sin{2021} > \frac{2\sin{1011}^2}{\sqrt{3}}
sin
1
+
sin
3
+
…
+
sin
2021
>
3
2
sin
1011
2
Proposed by Oleksii Masalitin
9.5
1
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King improves upon 8.4
Let
B
M
BM
BM
be the median of triangle
A
B
C
ABC
A
BC
in which
A
B
>
B
C
AB > BC
A
B
>
BC
. The point
P
P
P
is chosen so that
A
B
∥
P
C
AB\parallel PC
A
B
∥
PC
and
P
M
⊥
B
M
PM \perp BM
PM
⊥
BM
. On the line
B
P
BP
BP
, point
Q
Q
Q
is chosen so that
∠
A
Q
C
=
9
0
∘
\angle AQC = 90^\circ
∠
A
QC
=
9
0
∘
, and points
B
B
B
and
Q
Q
Q
are on opposite sides of the line
A
C
AC
A
C
. Prove that
A
B
=
B
Q
AB = BQ
A
B
=
BQ
.Proposed by Mykhailo Shtandenko
9.4
1
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Number of same colored numbers
You are given a positive integer
k
k
k
and not necessarily distinct positive integers
a
1
,
a
2
,
a
3
,
…
,
a
k
a_1, a_2 , a_3 , \ldots, a_k
a
1
,
a
2
,
a
3
,
…
,
a
k
. It turned out that for any coloring of all positive integers from
1
1
1
to
2021
2021
2021
in one of the
k
k
k
colors so that there are exactly
a
1
a_1
a
1
numbers of the first color,
a
2
a_2
a
2
numbers of the second color,
…
\ldots
…
, and
a
k
a_k
a
k
numbers of the
k
k
k
-th color, there is always a number
x
∈
{
1
,
2
,
…
,
2021
}
x \in \{1, 2, \ldots, 2021\}
x
∈
{
1
,
2
,
…
,
2021
}
, such that the total number of numbers colored in the same color as
x
x
x
is exactly
x
x
x
. What are the possible values of
k
k
k
?Proposed by Arsenii Nikolaiev
9.3
1
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Slowly growing product
Let
a
n
=
1
+
2
n
−
2
n
3
−
1
n
4
a_n = 1 + \frac{2}{n} - \frac{2}{n^3} - \frac{1}{n^4}
a
n
=
1
+
n
2
−
n
3
2
−
n
4
1
. For which smallest positive integer
n
n
n
does the value of
P
n
=
a
2
a
3
a
4
…
a
n
P_n = a_2a_3a_4 \ldots a_n
P
n
=
a
2
a
3
a
4
…
a
n
exceed
100
100
100
?
9.2
1
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Make almost equal
Roma wrote on the board each of the numbers
2018
,
2019
,
2020
2018, 2019, 2020
2018
,
2019
,
2020
,
100
100
100
times each. Let us denote by
S
(
n
)
S(n)
S
(
n
)
the sum of digits of positive integer
n
n
n
. In one action, Roma can choose any positive integer
k
k
k
and instead of any three numbers
a
,
b
,
c
a, b, c
a
,
b
,
c
written on the board write the numbers
2
S
(
a
+
b
)
+
k
,
2
S
(
b
+
c
)
+
k
2S(a + b) + k, 2S(b + c) + k
2
S
(
a
+
b
)
+
k
,
2
S
(
b
+
c
)
+
k
and
2
S
(
c
+
a
)
+
k
2S(c + a) + k
2
S
(
c
+
a
)
+
k
. Can Roma after several such actions make
299
299
299
numbers on the board equal, and the last one differing from them by
1
1
1
?Proposed by Oleksii Masalitin
9.1
1
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Birthday paradox
Before the math competition, Dmytro overheard Olena and Mykola talking about their birthdays.О: "The day and month of my birthday are half as large as the day and month of Mykola's birthday." М: "Also, the day of Olena's birth and the month of my birth are consecutive positive integers." О: "And the sum of all these four numbers is a multiple of
17
17
17
."Can Dmitro determine the day and month of Olena's birth?Proposed by Olena Artemchuk and Mykola Moroz
8.5
1
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Irreduced to ashes, modulo $p$
For a prime number
p
>
3
p > 3
p
>
3
, define the following irreducible fraction:
m
n
=
p
−
1
2
+
p
−
2
3
+
…
+
2
p
−
1
−
1
\frac{m}{n} = \frac{p-1}{2} + \frac{p-2}{3} + \ldots + \frac{2}{p-1} - 1
n
m
=
2
p
−
1
+
3
p
−
2
+
…
+
p
−
1
2
−
1
Prove that
m
m
m
is divisible by
p
p
p
.Proposed by Oleksii Masalitin
8.4
1
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Return of the geo king
Let
B
M
BM
BM
be the median of the triangle
A
B
C
ABC
A
BC
with
A
B
>
B
C
AB > BC
A
B
>
BC
. The point
P
P
P
is chosen so that
A
B
∥
P
C
AB\parallel PC
A
B
∥
PC
and
P
M
⊥
B
M
PM \perp BM
PM
⊥
BM
. Prove that
∠
A
B
M
=
∠
M
B
P
\angle ABM = \angle MBP
∠
A
BM
=
∠
MBP
.Proposed by Mykhailo Shandenko
8.3
1
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Almost magin square
The
1
×
1
1 \times 1
1
×
1
cells located around the perimeter of a
3
×
3
3 \times 3
3
×
3
square are filled with the numbers
1
,
2
,
…
,
8
1, 2, \ldots, 8
1
,
2
,
…
,
8
so that the sums along each of the four sides are equal. In the upper left corner cell is the number
8
8
8
, and in the upper left is the number
6
6
6
(see the figure below). https://i.ibb.co/bRmd12j/Kyiv-MO-2021-Round-1-8-2.png How many different ways to fill the remaining cells are there under these conditions?Proposed by Mariia Rozhkova
8.2
1
Hide problems
Erased digit
Oleksiy writes all the digits from
0
0
0
to
9
9
9
on the board, after which Vlada erases one of them. Then he writes
10
10
10
nine-digit numbers on the board, each consisting of all the nine digits written on the board (they don't have to be distinct). It turned out that the sum of these
10
10
10
numbers is a ten-digit number, all of whose digits are distinct. Which digit could have been erased by Vlada?Proposed by Oleksii Masalitin
8.1
1
Hide problems
Simple division trick
Find all positive integers
n
n
n
that can be subtracted from both the numerator and denominator of the fraction
1234
6789
\frac{1234}{6789}
6789
1234
, to get, after the reduction, the fraction of form
a
b
\frac{a}{b}
b
a
, where
a
,
b
a, b
a
,
b
are single digit numbers. Proposed by Bogdan Rublov
7.4
1
Hide problems
Magic rectangle (almost)
A rectangle
3
×
5
3 \times 5
3
×
5
is divided into
15
15
15
1
×
1
1 \times 1
1
×
1
cells. The middle
3
3
3
cells that have no common points with the border of the rectangle are deleted. Is it possible to put in the remaining
12
12
12
cells numbers
1
,
2
,
…
,
12
1, 2, \ldots, 12
1
,
2
,
…
,
12
in some order, so that the sums of the numbers in the cells along each of the four sides of the rectangle are equal?Proposed by Mariia Rozhkova
7.3
1
Hide problems
Small factoring
Petryk factored the number
1
0
6
=
1000000
10^6 = 1000000
1
0
6
=
1000000
as a product of
7
7
7
distinct positive integers. Among all such factorings, find the one in which the largest of these
7
7
7
factors is the smallest possible.Proposed by Bogdan Rublov
7.2
1
Hide problems
Cutting rectangle game
Andriy and Olesya take turns (Andriy starts) in a
2
×
1
2 \times 1
2
×
1
rectangle, drawing horizontal segments of length
2
2
2
or vertical segments of length
1
1
1
, as shown in the figure below. https://i.ibb.co/qWqWxgh/Kyiv-MO-2021-Round-1-7-2.pngAfter each move, the value
P
P
P
is calculated - the total perimeter of all small rectangles that are formed (i.e., those inside which no other segment passes). The winner is the one after whose move
P
P
P
is divisible by
2021
2021
2021
for the first time. Who has a winning strategy?Proposed by Bogdan Rublov
7.1
1
Hide problems
We balling
Mom brought Andriy and Olesya
4
4
4
balls with the numbers
1
,
2
,
3
1, 2, 3
1
,
2
,
3
and
4
4
4
written on them (one on each ball). She held
2
2
2
balls in each hand and did not know which numbers were written on the balls in each hand. The mother asked Andriy to take a ball with a higher number from each hand, and then to keep the ball with the lower number from the two balls he took. After that, she asked Olesya to take two other balls, and out of these two, keep the ball with the higher number. Does the mother know with certainty, which child has the ball with the higher number?Proposed by Bogdan Rublov