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Problems
Contests
International Contests
Romanian Masters of Mathematics Collection
2024 Romanian Master of Mathematics
2024 Romanian Master of Mathematics
Part of
Romanian Masters of Mathematics Collection
Subcontests
(6)
6
1
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Analytic NT on Polynomials with algebra assistance
A polynomial
P
P
P
with integer coefficients is square-free if it is not expressible in the form
P
=
Q
2
R
P = Q^2R
P
=
Q
2
R
, where
Q
Q
Q
and
R
R
R
are polynomials with integer coefficients and
Q
Q
Q
is not constant. For a positive integer
n
n
n
, let
P
n
P_n
P
n
be the set of polynomials of the form
1
+
a
1
x
+
a
2
x
2
+
⋯
+
a
n
x
n
1 + a_1x + a_2x^2 + \cdots + a_nx^n
1
+
a
1
x
+
a
2
x
2
+
⋯
+
a
n
x
n
with
a
1
,
a
2
,
…
,
a
n
∈
{
0
,
1
}
a_1,a_2,\ldots, a_n \in \{0,1\}
a
1
,
a
2
,
…
,
a
n
∈
{
0
,
1
}
. Prove that there exists an integer
N
N
N
such that for all integers
n
≥
N
n \geq N
n
≥
N
, more than
99
%
99\%
99%
of the polynomials in
P
n
P_n
P
n
are square-free.Navid Safaei, Iran
5
1
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Moving a point in the plane and magical fixed points
Let
B
C
BC
BC
be a fixed segment in the plane, and let
A
A
A
be a variable point in the plane not on the line
B
C
BC
BC
. Distinct points
X
X
X
and
Y
Y
Y
are chosen on the rays
C
A
→
CA^\to
C
A
→
and
B
A
→
BA^\to
B
A
→
, respectively, such that
∠
C
B
X
=
∠
Y
C
B
=
∠
B
A
C
\angle CBX = \angle YCB = \angle BAC
∠
CBX
=
∠
Y
CB
=
∠
B
A
C
. Assume that the tangents to the circumcircle of
A
B
C
ABC
A
BC
at
B
B
B
and
C
C
C
meet line
X
Y
XY
X
Y
at
P
P
P
and
Q
Q
Q
, respectively, such that the points
X
X
X
,
P
P
P
,
Y
Y
Y
and
Q
Q
Q
are pairwise distinct and lie on the same side of
B
C
BC
BC
. Let
Ω
1
\Omega_1
Ω
1
be the circle through
X
X
X
and
P
P
P
centred on
B
C
BC
BC
. Similarly, let
Ω
2
\Omega_2
Ω
2
be the circle through
Y
Y
Y
and
Q
Q
Q
centred on
B
C
BC
BC
. Prove that
Ω
1
\Omega_1
Ω
1
and
Ω
2
\Omega_2
Ω
2
intersect at two fixed points as
A
A
A
varies.Daniel Pham Nguyen, Denmark
4
1
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Innovative or not so much NT with powers and remainders
Fix integers
a
a
a
and
b
b
b
greater than
1
1
1
. For any positive integer
n
n
n
, let
r
n
r_n
r
n
be the (non-negative) remainder that
b
n
b^n
b
n
leaves upon division by
a
n
a^n
a
n
. Assume there exists a positive integer
N
N
N
such that
r
n
<
2
n
n
r_n < \frac{2^n}{n}
r
n
<
n
2
n
for all integers
n
≥
N
n\geq N
n
≥
N
. Prove that
a
a
a
divides
b
b
b
.Pouria Mahmoudkhan Shirazi, Iran
3
1
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From combinatorics to geometry, through algebra
Given a positive integer
n
n
n
, a collection
S
\mathcal{S}
S
of
n
−
2
n-2
n
−
2
unordered triples of integers in
{
1
,
2
,
…
,
n
}
\{1,2,\ldots,n\}
{
1
,
2
,
…
,
n
}
is
n
n
n
-admissible if for each
1
≤
k
≤
n
−
2
1 \leq k \leq n - 2
1
≤
k
≤
n
−
2
and each choice of
k
k
k
distinct
A
1
,
A
2
,
…
,
A
k
∈
S
A_1, A_2, \ldots, A_k \in \mathcal{S}
A
1
,
A
2
,
…
,
A
k
∈
S
we have
∣
A
1
∪
A
2
∪
⋯
A
k
∣
≥
k
+
2.
\left|A_1 \cup A_2 \cup \cdots A_k \right| \geq k+2.
∣
A
1
∪
A
2
∪
⋯
A
k
∣
≥
k
+
2.
Is it true that for all
n
>
3
n > 3
n
>
3
and for each
n
n
n
-admissible collection
S
\mathcal{S}
S
, there exist pairwise distinct points
P
1
,
…
,
P
n
P_1, \ldots , P_n
P
1
,
…
,
P
n
in the plane such that the angles of the triangle
P
i
P
j
P
k
P_iP_jP_k
P
i
P
j
P
k
are all less than
6
1
∘
61^{\circ}
6
1
∘
for any triple
{
i
,
j
,
k
}
\{i, j, k\}
{
i
,
j
,
k
}
in
S
\mathcal{S}
S
?Ivan Frolov, Russia
2
1
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Combinatorics on partial sums and divisibility
Consider an odd prime
p
p
p
and a positive integer
N
<
50
p
N < 50p
N
<
50
p
. Let
a
1
,
a
2
,
…
,
a
N
a_1, a_2, \ldots , a_N
a
1
,
a
2
,
…
,
a
N
be a list of positive integers less than
p
p
p
such that any specific value occurs at most
51
100
N
\frac{51}{100}N
100
51
N
times and
a
1
+
a
2
+
⋯
⋅
+
a
N
a_1 + a_2 + \cdots· + a_N
a
1
+
a
2
+
⋯
⋅
+
a
N
is not divisible by
p
p
p
. Prove that there exists a permutation
b
1
,
b
2
,
…
,
b
N
b_1, b_2, \ldots , b_N
b
1
,
b
2
,
…
,
b
N
of the
a
i
a_i
a
i
such that, for all
k
=
1
,
2
,
…
,
N
k = 1, 2, \ldots , N
k
=
1
,
2
,
…
,
N
, the sum
b
1
+
b
2
+
⋯
+
b
k
b_1 + b_2 + \cdots + b_k
b
1
+
b
2
+
⋯
+
b
k
is not divisible by
p
p
p
.Will Steinberg, United Kingdom
1
1
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Bishops and permutations
Let
n
n
n
be a positive integer. Initially, a bishop is placed in each square of the top row of a
2
n
×
2
n
2^n \times 2^n
2
n
×
2
n
chessboard; those bishops are numbered from
1
1
1
to
2
n
2^n
2
n
from left to right. A jump is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.Find the total number of permutations
σ
\sigma
σ
of the numbers
1
,
2
,
…
,
2
n
1, 2, \ldots, 2^n
1
,
2
,
…
,
2
n
with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order
σ
(
1
)
,
σ
(
2
)
,
…
,
σ
(
2
n
)
\sigma(1), \sigma(2), \ldots, \sigma(2^n)
σ
(
1
)
,
σ
(
2
)
,
…
,
σ
(
2
n
)
, from left to right.Israel