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Problems
Contests
National and Regional Contests
Argentina Contests
Argentina National Olympiad
2002 Argentina National Olympiad
2002 Argentina National Olympiad
Part of
Argentina National Olympiad
Subcontests
(6)
6
1
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n infinite arithmetic progressions of positive integers
Let
P
1
,
P
2
,
…
,
P
n
P_1,P_2,\ldots ,P_n
P
1
,
P
2
,
…
,
P
n
, be infinite arithmetic progressions of positive integers, of differences
d
1
,
d
2
,
…
,
d
n
d_1,d_2,\ldots ,d_n
d
1
,
d
2
,
…
,
d
n
, respectively. Prove that if every positive integer appears in at least one of the
n
n
n
progressions then one of the differences
d
i
d_i
d
i
divides the least common multiple of the remaining
n
−
1
n-1
n
−
1
differences. Note:
P
i
=
{
a
i
,
a
i
+
d
i
,
a
i
+
2
d
i
,
a
i
+
3
d
i
,
a
i
+
4
d
i
,
⋯
}
P_i=\left \{ a_i,a_i+d_i,a_i+2d_i,a_i+3d_i,a_i+4d_i,\cdots \right \}
P
i
=
{
a
i
,
a
i
+
d
i
,
a
i
+
2
d
i
,
a
i
+
3
d
i
,
a
i
+
4
d
i
,
⋯
}
with
a
i
a_i
a
i
and
d
i
d_i
d
i
positive integers.
4
1
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after completing 2001 steps, only one number remains on the board.
Initially on the blackboard all the integers from
1
1
1
to
2002
2002
2002
inclusive are written in one line and in some order, without repetitions. In each step, the first and second numbers of the line are deleted and the absolute value of the subtraction of the two numbers that have just been deleted is written at the beginning of the line; the other numbers are not modified in that step, and there is a new line that has one less number than the previous step. After completing
2001
2001
2001
steps, only one number remains on the board. Determine all possible values of the number left on the board by varying the order of the
2002
2002
2002
numbers on the initial line (and performing the
2001
2001
2001
steps).
1
1
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sum of numbers of fibanacci like sequence
On the computer screen there are initially two
1
1
1
's written. The insert program causes the sum of those numbers to be inserted between each pair of numbers by pressing the
E
n
t
e
r
Enter
E
n
t
er
key. In the first step a number is inserted and we obtain
1
−
2
−
1
1-2-1
1
−
2
−
1
; In the second step two numbers are inserted and we have
1
−
3
−
2
−
3
−
1
1-3-2-3-1
1
−
3
−
2
−
3
−
1
; In the third, four numbers are inserted and you have
1
−
4
−
3
−
5
−
2
−
5
−
3
−
4
−
1
1-4-3-5-2-5-3-4-1
1
−
4
−
3
−
5
−
2
−
5
−
3
−
4
−
1
; etc Find the sum of all the numbers that appear on the screen at the end of step number
25
25
25
.
2
1
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2002x+273y=200201+k diophantine
Determine the smallest positive integer
k
k
k
so that the equation
2002
x
+
273
y
=
200201
+
k
2002x+273y=200201+k
2002
x
+
273
y
=
200201
+
k
has integer solutions, and for that value of
k
k
k
, find the number of solutions
(
x
,
y
)
\left (x,y\right )
(
x
,
y
)
with
x
x
x
,
y
y
y
positive integers that have the equation.
3
1
Hide problems
4 circles, tangent in many ways, r_1+r_2+r_3=r
In a circumference
Γ
\Gamma
Γ
a chord
P
Q
PQ
PQ
is considered such that the segment that joins the midpoint of the smallest arc
P
Q
PQ
PQ
and the midpoint of the segment
P
Q
PQ
PQ
measures
1
1
1
. Let
Γ
1
,
Γ
2
\Gamma_1, \Gamma_2
Γ
1
,
Γ
2
and
Γ
3
\Gamma_3
Γ
3
be three tangent circumferences to the chord
P
Q
PQ
PQ
that are in the same half plane than the center of
Γ
\Gamma
Γ
with respect to the line
P
Q
PQ
PQ
. Furthermore,
Γ
1
\Gamma_1
Γ
1
and
Γ
3
\Gamma_3
Γ
3
are internally tangent to
Γ
\Gamma
Γ
and externally tangent to
Γ
2
\Gamma_2
Γ
2
, and the centers of
Γ
1
\Gamma_1
Γ
1
and
Γ
3
\Gamma_3
Γ
3
are on different halfplanes with respect to the line determined by the centers of
Γ
\Gamma
Γ
and
Γ
2
\Gamma_2
Γ
2
. If the sum of the radii of
Γ
1
,
Γ
2
\Gamma_1, \Gamma_2
Γ
1
,
Γ
2
and
Γ
3
\Gamma_3
Γ
3
is equal to the radius of
Γ
\Gamma
Γ
, calculate the radius of
Γ
2
\Gamma_2
Γ
2
.
5
1
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perpendicularity wanted, isosceles triangle and parallelogram related
Let
△
A
B
C
\vartriangle ABC
△
A
BC
be an isosceles triangle with
A
C
=
B
C
AC = BC
A
C
=
BC
. Points
D
,
E
,
F
D, E, F
D
,
E
,
F
are considered on
B
C
,
C
A
,
A
B
BC, CA, AB
BC
,
C
A
,
A
B
, respectively, such that
A
F
>
B
F
AF> BF
A
F
>
BF
and that the quadrilateral
C
E
F
D
CEFD
CEF
D
is a parallelogram. The perpendicular line to
B
C
BC
BC
drawn by
B
B
B
intersects the perpendicular bisector of
A
B
AB
A
B
at
G
G
G
. Prove that
D
E
⊥
F
G
DE \perp FG
D
E
⊥
FG
.