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Problems
Contests
National and Regional Contests
Serbia Contests
Serbia Team Selection Test
2017 Serbia Team Selection Test
2017 Serbia Team Selection Test
Part of
Serbia Team Selection Test
Subcontests
(6)
6
1
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Intersting divisor number theory
Let
k
k
k
be a positive integer and let
n
n
n
be the smallest number with exactly
k
k
k
divisors. Given
n
n
n
is a cube, is it possible that
k
k
k
is divisible by a prime factor of the form
3
j
+
2
3j+2
3
j
+
2
?
4
1
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Covering unit segments
We have an
n
×
n
n \times n
n
×
n
square divided into unit squares. Each side of unit square is called unit segment. Some isoceles right triangles of hypotenuse
2
2
2
are put on the square so all their vertices are also vertices of unit squares. For which
n
n
n
it is possible that every unit segment belongs to exactly one triangle(unit segment belongs to a triangle even if it's on the border of the triangle)?
5
1
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A set of inequalities on a sequence [Serbia TST 2017, D2, P2]
Let
n
≥
2
n \geq 2
n
≥
2
be a positive integer and
{
x
i
}
i
=
0
n
\{x_i\}_{i=0}^n
{
x
i
}
i
=
0
n
a sequence such that not all of its elements are zero and there is a positive constant
C
n
C_n
C
n
for which: (i)
x
1
+
⋯
+
x
n
=
0
x_1+ \dots +x_n=0
x
1
+
⋯
+
x
n
=
0
, and (ii) for each
i
i
i
either
x
i
≤
x
i
+
1
x_i\leq x_{i+1}
x
i
≤
x
i
+
1
or
x
i
≤
x
i
+
1
+
C
n
x
i
+
2
x_i\leq x_{i+1} + C_n x_{i+2}
x
i
≤
x
i
+
1
+
C
n
x
i
+
2
(all indexes are assumed modulo
n
n
n
). Prove that a)
C
n
≥
2
C_n\geq 2
C
n
≥
2
, and b)
C
n
=
2
C_n=2
C
n
=
2
if and only
2
∣
n
2 \mid n
2
∣
n
.
1
1
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Non-stanard easy geo inequality [Serbia TST 2017, D1, P1]
Let
A
B
C
ABC
A
BC
be a triangle and
D
D
D
the midpoint of the side
B
C
BC
BC
. Define points
E
E
E
and
F
F
F
on
A
C
AC
A
C
and
B
B
B
, respectively, such that
D
E
=
D
F
DE=DF
D
E
=
D
F
and
∠
E
D
F
=
∠
B
A
C
\angle EDF =\angle BAC
∠
E
D
F
=
∠
B
A
C
. Prove that
D
E
≥
A
B
+
A
C
4
.
DE\geq \frac {AB+AC} 4.
D
E
≥
4
A
B
+
A
C
.
2
1
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Interesting process
Initally a pair
(
x
,
y
)
(x, y)
(
x
,
y
)
is written on the board, such that exactly one of it's coordinates is odd. On such a pair we perform an operation to get pair
(
x
2
,
y
+
x
2
)
(\frac x 2, y+\frac x 2)
(
2
x
,
y
+
2
x
)
if
2
∣
x
2|x
2∣
x
and
(
x
+
y
2
,
y
2
)
(x+\frac y 2, \frac y 2)
(
x
+
2
y
,
2
y
)
if
2
∣
y
2|y
2∣
y
. Prove that for every odd
n
>
1
n>1
n
>
1
there is a even positive integer
b
<
n
b<n
b
<
n
such that starting from the pair
(
n
,
b
)
(n, b)
(
n
,
b
)
we will get the pair
(
b
,
n
)
(b, n)
(
b
,
n
)
after finitely many operations.
3
1
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Integer function
A function
f
:
N
→
N
f:\mathbb{N} \rightarrow \mathbb{N}
f
:
N
→
N
is called nice if
f
a
(
b
)
=
f
(
a
+
b
−
1
)
f^a(b)=f(a+b-1)
f
a
(
b
)
=
f
(
a
+
b
−
1
)
, where
f
a
(
b
)
f^a(b)
f
a
(
b
)
denotes
a
a
a
times applied function
f
f
f
. Let
g
g
g
be a nice function, and an integer
A
A
A
exists such that
g
(
A
+
2018
)
=
g
(
A
)
+
1
g(A+2018)=g(A)+1
g
(
A
+
2018
)
=
g
(
A
)
+
1
. a) Prove that
g
(
n
+
201
7
2017
)
=
g
(
n
)
g(n+2017^{2017})=g(n)
g
(
n
+
201
7
2017
)
=
g
(
n
)
for all
n
≥
A
+
2
n \geq A+2
n
≥
A
+
2
. b) If
g
(
A
+
1
)
≠
g
(
A
+
1
+
201
7
2017
)
g(A+1) \neq g(A+1+2017^{2017})
g
(
A
+
1
)
=
g
(
A
+
1
+
201
7
2017
)
find
g
(
n
)
g(n)
g
(
n
)
for
n
<
A
n <A
n
<
A
.