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Problems
Contests
National and Regional Contests
Argentina Contests
Argentina National Olympiad
2009 Argentina National Olympiad
2009 Argentina National Olympiad
Part of
Argentina National Olympiad
Subcontests
(6)
6
1
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a_{n+1}=m a_n where m is integer , S_n= the sum of digits of a_n
A sequence
a
0
,
a
1
,
a
2
,
.
.
.
,
a
n
,
.
.
.
a_0,a_1,a_2,...,a_n,...
a
0
,
a
1
,
a
2
,
...
,
a
n
,
...
is such that
a
0
=
1
a_0=1
a
0
=
1
and, for each
n
≥
0
n\ge 0
n
≥
0
,
a
n
+
1
=
m
⋅
a
n
a_{n+1}=m \cdot a_n
a
n
+
1
=
m
⋅
a
n
, where
m
m
m
is an integer between
2
2
2
and
9
9
9
inclusive. Also, every integer between
2
2
2
and
9
9
9
has even been used at least once to get
a
n
+
1
a_{n+1}
a
n
+
1
from
a
n
a_n
a
n
. Let
S
n
Sn
S
n
the sum of the digits of
a
n
a_n
a
n
,
n
=
0
,
1
,
2
,
.
.
.
n=0,1,2,...
n
=
0
,
1
,
2
,
...
. Prove that
S
n
≥
S
n
+
1
S_n \ge S_{n+1}
S
n
≥
S
n
+
1
for infinite values of
n
n
n
.
5
1
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2009 integers around a circle
Around a circle are written
2009
2009
2009
integers, not necessarily distinct, so that if two numbers are neighbors their difference is
1
1
1
or
2
2
2
. We will say that a number is huge if it is greater than its two neighbors, and that it is tiny if it is less than its two neighbors. The sum of all the huge numbers is equal to the sum of all the tiny numbers plus
1810
1810
1810
. . Determine how many odd numbers there can be around the circumference.
4
1
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100 equal rods
You have
100
100
100
equal rods. It is allowed to split each rod into two or three shorter rods, not necessarily the same. The objective is that by rearranging the pieces (and using them all)
q
>
200
q>200
q
>
200
can be assembled new rods, all of equal length. Find the values of
q
q
q
for whom this can be done.
2
1
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sum of squares of its proper divisors is equal to 2n+4
A positive integer
n
n
n
is acceptable if the sum of the squares of its proper divisors is equal to
2
n
+
4
2n+4
2
n
+
4
(a divisor of
n
n
n
is proper if it is different from
1
1
1
and of
n
n
n
). Find all acceptable numbers less than
10000
10000
10000
,
1
1
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2009 points on circle, 7 colors
2009
2009
2009
points have been marked on a circle. Lucía colors them with
7
7
7
different colors of her choice. Then Ivan can join three points of the same color, thus forming monochrome triangles. Triangles cannot have points in common; not even vertices in common. Ivan's goal is to draw as many monochrome triangles as possible. Lucía's objective is to prevent Iván's task as much as possible through a good choice of colouring. How many monochrome triangles will Ivan get if they both do their homework to the best of their ability?
3
1
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Isosceles trapezoid
Isosceles trapezoid
A
B
C
D
ABCD
A
BC
D
, with
A
B
∥
C
D
AB \parallel CD
A
B
∥
C
D
, is such that there exists a circle
Γ
\Gamma
Γ
tangent to its four sides. Let T \equal{} \Gamma \cap BC, and P \equal{} \Gamma \cap AT (
P
≠
T
P \neq T
P
=
T
). If \frac{AP}{AT} \equal{} \frac{2}{5}, compute
A
B
C
D
\frac{AB}{CD}
C
D
A
B
.