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Problems
Contests
National and Regional Contests
Germany Contests
QEDMO
2012 QEDMO 11th
2012 QEDMO 11th
Part of
QEDMO
Subcontests
(11)
3
1
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2^n species on planet Kerbin
Today there are
2
n
2^n
2
n
species on the planet Kerbin, all of which are exactly n steps from an original species. In an evolutionary step, One species split into exactly two new species and died out in the process. There were already
2
n
−
1
2^n-1
2
n
−
1
species in the past, which are no longer present today can be found, but are only documented by fossils. The famous space pioneer Jebediah Kerman once suggested reducing the biodiversity of a planet by doing this to measure how closely two species are on average related, with also already extinct species should be taken into account. The degree of relationship is measured two types, of course, by how many evolutionary steps before or back you have to do at least one to get from one to the other. What is the biodiversity of the planet Kerbin?
7
1
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rhombuses cover a hexagon into 3 orientations
In the following, a rhombus is one with edge length
1
1
1
and interior angles
6
0
o
60^o
6
0
o
and
12
0
o
120^o
12
0
o
. Now let
n
n
n
be a natural number and
H
H
H
a regular hexagon with edge length
n
n
n
, which is covered with rhombuses without overlapping has been. The rhombuses then appear in three different orientations. Prove that whatever the overlap looks exactly, each of these three orientations can be viewed at the same time.
10
1
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3 cups with 3 interger units of gallium
Let there be three cups
A
,
B
A, B
A
,
B
and
C
C
C
, which start with
a
,
b
a, b
a
,
b
and
c
c
c
(all of them are natural numbers) units of gallium filled. It is also believed that all cups are large enough to contain the total amount of gallium available. It is now allowed to move gallium from one cup to another cup, provided that the contents of the latter cup are exactly double. (a) For which starting positions is it possible to empty one of the cups? (b) For which starting positions is it possible to put all of the gallium in one cup?
5
1
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n coloured lattice points , same color with ther center of gravity, bw
Let
n
n
n
be a natural number and
L
=
Z
2
L = Z^2
L
=
Z
2
the set of points on the plane with integer coordinates. Every point in
L
L
L
is colored now in one of the colors red or green. Show that there are
n
n
n
different points
x
1
,
.
.
.
,
x
n
∈
L
x_1,...,x_n \in L
x
1
,
...
,
x
n
∈
L
all of which have the same color and whose center of gravity is also in
L
L
L
and is of the same color.
6
1
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1^{p-1} + 2^{p-1} +...+ (p-1)^{p-1} = p + (p-1)! \mod p^2
Let
p
p
p
be an odd prime number. Prove that
1
p
−
1
+
2
p
−
1
+
.
.
.
+
(
p
−
1
)
p
−
1
≡
p
+
(
p
−
1
)
!
m
o
d
p
2
1^{p-1} + 2^{p-1} +...+ (p-1)^{p-1} \equiv p + (p-1)! \mod p^2
1
p
−
1
+
2
p
−
1
+
...
+
(
p
−
1
)
p
−
1
≡
p
+
(
p
−
1
)!
mod
p
2
8
1
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2012 points with integer distances in pairs
Prove that there are
2012
2012
2012
points in the plane, none of which are three on one straight line and in pairs have integer distances .
11
1
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f (xf (y) + f (x)) = xy$
Find all functions
f
:
R
→
R
f: R\to R
f
:
R
→
R
, such that
f
(
x
f
(
y
)
+
f
(
x
)
)
=
x
y
f (xf (y) + f (x)) = xy
f
(
x
f
(
y
)
+
f
(
x
))
=
x
y
for all
x
,
y
∈
R
x, y \in R
x
,
y
∈
R
.
12
1
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infinite naturals in form k^2 + 1 without any real divisors of same form
Prove that there are infinitely many different natural numbers of the form
k
2
+
1
k^2 + 1
k
2
+
1
,
k
∈
N
k \in N
k
∈
N
that have no real divisor of this form.
4
1
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bw in nxn chessboard so that in every 2x2 both colors have same number
The fields of an
n
×
n
n\times n
n
×
n
chess board are colored black and white, such that in every small
2
×
2
2\times 2
2
×
2
-square both colors should be the same number. How many there possibilities are for this?
2
1
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heads and tails with n unfair coins
N
N
N
unfair coins (with heads and tails on the sides) are thrown, with the
k
t
h
k^{th}
k
t
h
coin has got a chance of
1
2
k
+
1
\frac{1}{2k + 1}
2
k
+
1
1
to land on tails.How high is the probability that an odd number of coins will show tails?
1
1
Hide problems
(2^x + 1) (2^y-1) = 2^z-1
Find all
x
,
y
,
z
∈
N
0
x, y, z \in N_0
x
,
y
,
z
∈
N
0
with
(
2
x
+
1
)
(
2
y
−
1
)
=
2
z
−
1
(2^x + 1) (2^y-1) = 2^z-1
(
2
x
+
1
)
(
2
y
−
1
)
=
2
z
−
1
.