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National and Regional Contests
Serbia Contests
Serbia JBMO TST
2022 Serbia JBMO TST
2022 Serbia JBMO TST
Part of
Serbia JBMO TST
Subcontests
(4)
4
1
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5x5 board with n in all cells
Initially in every cell of a
5
×
5
5\times 5
5
×
5
board is the number
0
0
0
. In one move you may take any cell of this board and add
1
1
1
to it and all of its adjacent cells (two cells are adjacent if they share an edge). After a finite number of moves, number
n
n
n
is written in all cells. Find all possible values of
n
n
n
.
3
1
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5 conditions
Find all natural numbers
n
n
n
for which the following
5
5
5
conditions hold:
(
1
)
(1)
(
1
)
n
n
n
is not divisible by any perfect square bigger than
1
1
1
.
(
2
)
(2)
(
2
)
n
n
n
has exactly one prime divisor of the form
4
k
+
3
4k+3
4
k
+
3
,
k
∈
N
0
k\in \mathbb{N}_0
k
∈
N
0
.
(
3
)
(3)
(
3
)
Denote by
S
(
n
)
S(n)
S
(
n
)
the sum of digits of
n
n
n
and
d
(
n
)
d(n)
d
(
n
)
as the number of positive divisors of
n
n
n
. Then we have that
S
(
n
)
+
2
=
d
(
n
)
S(n)+2=d(n)
S
(
n
)
+
2
=
d
(
n
)
.
(
4
)
(4)
(
4
)
n
+
3
n+3
n
+
3
is a perfect square.
(
5
)
(5)
(
5
)
n
n
n
does not have a prime divisor which has
4
4
4
or more digits.
2
1
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Collinear points, prove 90 angle
Let
I
I
I
be the incenter,
A
1
A_1
A
1
and
B
1
B_1
B
1
midpoints of sides
B
C
BC
BC
and
A
C
AC
A
C
of a triangle
Δ
A
B
C
\Delta ABC
Δ
A
BC
. Denote by
M
M
M
and
N
N
N
the midpoints of the arcs
A
C
AC
A
C
and
B
C
BC
BC
of circumcircle of
Δ
A
B
C
\Delta ABC
Δ
A
BC
which do contain the other vertex of the triangle. If points
M
M
M
,
I
I
I
and
N
N
N
are collinear prove that: \begin{align*} \angle AIB_1=\angle BIA_1=90^{\circ} \end{align*}
1
1
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Inequality with mathematical means
Prove that for all positive real numbers
a
a
a
,
b
b
b
the following inequality holds: \begin{align*} \sqrt{\frac{a^2+b^2}{2}}+\frac{2ab}{a+b}\ge \frac{a+b}{2}+ \sqrt{ab} \end{align*} When does equality hold?