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Problems
Contests
National and Regional Contests
Sweden Contests
Swedish Mathematical Competition
2020 Swedish Mathematical Competition
2020 Swedish Mathematical Competition
Part of
Swedish Mathematical Competition
Subcontests
(6)
6
1
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axis parallel cubes combo geo
A finite set of axis parallel cubes in space has the property of each point of the room is located in a maximum of M different cubes. Show that you can divide the amount of cubes in
8
(
M
−
1
)
+
1
8 (M - 1) + 1
8
(
M
−
1
)
+
1
subsets (or less) with the property that the cubes in each subset lacks common points. (An axis parallel cube is a cube whose edges are parallel to the coordinate axes.)
5
1
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p divides sum of combinations (p-1,k)a^k
Find all integers
a
a
a
such that there is a prime number of
p
≥
5
p\ge 5
p
≥
5
that divides
(
p
−
1
2
)
{p-1 \choose 2}
(
2
p
−
1
)
+
(
p
−
1
3
)
a
+ {p-1 \choose 3} a
+
(
3
p
−
1
)
a
+
(
p
−
1
4
)
a
2
+{p-1 \choose 4} a^2
+
(
4
p
−
1
)
a
2
+ ...+
(
p
−
1
p
−
3
)
a
p
−
5
.
{p-1 \choose p-3} a^{p-5} .
(
p
−
3
p
−
1
)
a
p
−
5
.
4
1
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n-gon with sides 1,2,3,.., n and vertices at lattice points
Which is the least positive integer
n
n
n
for which it is possible to find a (non-degenerate)
n
n
n
-gon with sidelengths
1
,
2
,
.
.
.
,
n
1, 2,. . . , n
1
,
2
,
...
,
n
, and where all vertices have integer coordinates?
3
1
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f (f (x) + y) = f (x) + f (y), bounded
Determine all bounded functions
f
:
R
→
R
f: R \to R
f
:
R
→
R
, such that
f
(
f
(
x
)
+
y
)
=
f
(
x
)
+
f
(
y
)
f (f (x) + y) = f (x) + f (y)
f
(
f
(
x
)
+
y
)
=
f
(
x
)
+
f
(
y
)
, for all real
x
,
y
x, y
x
,
y
.
1
1
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numbers of 1x2x3, 2x 3x4,..., 2020x 2021x2022 divisible by 2020
How many of the numbers
1
⋅
2
⋅
3
1\cdot 2\cdot 3
1
⋅
2
⋅
3
,
2
⋅
3
⋅
4
2\cdot 3\cdot 4
2
⋅
3
⋅
4
,...,
2020
⋅
2021
⋅
2022
2020 \cdot 2021 \cdot 2022
2020
⋅
2021
⋅
2022
are divisible by
2020
2020
2020
?
2
1
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1/2 < AC/ BC < 2 for perpendicular medians
The medians of the sides
A
C
AC
A
C
and
B
C
BC
BC
in the triangle
A
B
C
ABC
A
BC
are perpendicular to each other. Prove that
1
2
<
∣
A
C
∣
∣
B
C
∣
<
2
\frac12 <\frac{|AC|}{|BC|}<2
2
1
<
∣
BC
∣
∣
A
C
∣
<
2
.