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Portugal Contests
Portugal MO
2000 Portugal MO
2000 Portugal MO
Part of
Portugal MO
Subcontests
(6)
1
1
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change numbers in 3x3 table - Portugal OPM 2000 p1
Consider the following table where initially all squares contain zeros: \begin{tabular}{ | l | c | r| } \hline 0 & 0 & 0 \\ \hline 0 & 0 & 0 \\ \hline 0 & 0 & 0 \\ \hline \end{tabular} To change the table, the following operation is allowed: a
2
×
2
2 \times 2
2
×
2
square formed by adjacent squares is chosen, and a unit is added to all its numbers. Complete the following table, knowing that it was obtained by a sequence of permitted operations \begin{tabular}{ | l | c | r| } \hline 14 & & \\ \hline 19 & 36 & \\ \hline & 16 & \\ \hline \end{tabular}
6
1
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n players in a tournament, there can be only one champion
In a tournament,
n
n
n
players participate. Each player plays each other exactly once, with no ties. A player
A
A
A
is said to be champion if, for every other player
B
B
B
, one of the following two situations occurs: (a)
A
A
A
beat
B
B
B
; (b)
A
A
A
beat a player
C
C
C
who in turn beat
B
B
B
. Prove that in such a tournament there cannot be exactly two champions.
4
1
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sum of all numbers with odd digits once - Portugal OPM 2000 p4
Calculates the sum of all numbers that can be formed using each of the odd digits once, that is, the numbers
13579
13579
13579
,
13597
13597
13597
, ...,
97531
97531
97531
.
3
1
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max k, 2^k is a divisor of 3^n+1
Determine, for each positive integer
n
n
n
, the largest positive integer
k
k
k
such that
2
k
2^k
2
k
is a divisor of
3
n
+
1
3^n+1
3
n
+
1
.
5
1
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computation, right triangle inside right triangle 2000 Portugal p5
In the figure,
[
A
B
C
]
[ABC]
[
A
BC
]
and
[
D
E
C
]
[DEC]
[
D
EC
]
are right triangles . Knowing that
E
B
=
1
/
2
,
E
C
=
1
EB = 1/2, EC = 1
EB
=
1/2
,
EC
=
1
and
A
D
=
1
AD = 1
A
D
=
1
, calculate
D
C
DC
D
C
. https://1.bp.blogspot.com/-nAOrVnK5JmI/X4UMb2CNTyI/AAAAAAAAMmk/TtaBESxYyJ0FsBoY2XaCGlCTc6mgmA5TQCLcBGAsYHQ/s0/2000%2Bportugal%2Bp5.png
2
1
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computational geometry, 2-digit lengths, rational 2000 Portugal p2
In the figure, the chord
[
C
D
]
[CD]
[
C
D
]
is perpendicular to the diameter
[
A
B
]
[AB]
[
A
B
]
and intersects it at
H
H
H
. Length of
A
B
AB
A
B
is a two-digit natural number. Changing the order of these two digits gives length of
C
D
CD
C
D
. Knowing that distance from
H
H
H
to the center
O
O
O
is a positive rational number, calculate
A
B
AB
A
B
. https://cdn.artofproblemsolving.com/attachments/5/f/eb9c61579a38118b4f753bbc19a9a50e0732dc.png