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Contests
National and Regional Contests
Romania Contests
Romania Team Selection Test
2011 Romania Team Selection Test
2011 Romania Team Selection Test
Part of
Romania Team Selection Test
Subcontests
(4)
2
3
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arithmetic progressions
Prove that the set
S
=
{
⌊
n
π
⌋
∣
n
=
0
,
1
,
2
,
3
,
…
}
S=\{\lfloor n\pi\rfloor \mid n=0,1,2,3,\ldots\}
S
=
{⌊
nπ
⌋
∣
n
=
0
,
1
,
2
,
3
,
…
}
contains arithmetic progressions of any finite length, but no infinite arithmetic progressions.Vasile Pop
Concurrent on the Euler line
In triangle
A
B
C
ABC
A
BC
, the incircle touches sides
B
C
,
C
A
BC,CA
BC
,
C
A
and
A
B
AB
A
B
in
D
,
E
D,E
D
,
E
and
F
F
F
respectively. Let
X
X
X
be the feet of the altitude of the vertex
D
D
D
on side
E
F
EF
EF
of triangle
D
E
F
DEF
D
EF
. Prove that
A
X
,
B
Y
AX,BY
A
X
,
B
Y
and
C
Z
CZ
CZ
are concurrent on the Euler line of the triangle
D
E
F
DEF
D
EF
.
A matrix of 0s and 1s
Given a prime number
p
p
p
congruent to
1
1
1
modulo
5
5
5
such that
2
p
+
1
2p+1
2
p
+
1
is also prime, show that there exists a matrix of
0
0
0
s and
1
1
1
s containing exactly
4
p
4p
4
p
(respectively,
4
p
+
2
4p+2
4
p
+
2
)
1
1
1
s no sub-matrix of which contains exactly
2
p
2p
2
p
(respectively,
2
p
+
1
2p+1
2
p
+
1
)
1
1
1
s.
3
4
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4
2
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disjoint cycles
Given an integer
n
≥
2
n\ge 2
n
≥
2
, compute
∑
σ
sgn
(
σ
)
n
ℓ
(
σ
)
\sum_{\sigma} \textrm{sgn}(\sigma) n^{\ell(\sigma)}
∑
σ
sgn
(
σ
)
n
ℓ
(
σ
)
, where all
n
n
n
-element permutations are considered, and where
ℓ
(
σ
)
\ell(\sigma)
ℓ
(
σ
)
is the number of disjoint cycles in the standard decomposition of
σ
\sigma
σ
.
Sum of consecutive squares
Show that: a) There are infinitely many positive integers
n
n
n
such that there exists a square equal to the sum of the squares of
n
n
n
consecutive positive integers (for instance,
2
2
2
is one such number as
5
2
=
3
2
+
4
2
5^2=3^2+4^2
5
2
=
3
2
+
4
2
). b) If
n
n
n
is a positive integer which is not a perfect square, and if
x
x
x
is an integer number such that
x
2
+
(
x
+
1
)
2
+
.
.
.
+
(
x
+
n
−
1
)
2
x^2+(x+1)^2+...+(x+n-1)^2
x
2
+
(
x
+
1
)
2
+
...
+
(
x
+
n
−
1
)
2
is a perfect square, then there are infinitely many positive integers
y
y
y
such that
y
2
+
(
y
+
1
)
2
+
.
.
.
+
(
y
+
n
−
1
)
2
y^2+(y+1)^2+...+(y+n-1)^2
y
2
+
(
y
+
1
)
2
+
...
+
(
y
+
n
−
1
)
2
is a perfect square.
1
5
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