MathDB
Problems
Contests
National and Regional Contests
Bosnia Herzegovina Contests
Regional Olympiad - Republic of Srpska
2004 Regional Olympiad - Republic of Srpska
2004 Regional Olympiad - Republic of Srpska
Part of
Regional Olympiad - Republic of Srpska
Subcontests
(4)
4
3
Hide problems
Regional Olympiad - Republic of Srpska 2004 Grade 9 Problem 4
Set
S
=
{
1
,
2
,
.
.
.
,
n
}
S=\{1,2,...,n\}
S
=
{
1
,
2
,
...
,
n
}
is firstly divided on
m
m
m
disjoint nonempty subsets, and then on
m
2
m^2
m
2
disjoint nonempty subsets. Prove that some
m
m
m
elements of set
S
S
S
were after first division in same set, and after the second division were in
m
m
m
different sets
n-gon and congruences
A convex
n
n
n
-gon
A
1
A
2
…
A
n
A_1A_2\dots A_n
A
1
A
2
…
A
n
(
n
>
3
)
(n>3)
(
n
>
3
)
is divided into triangles by non-intersecting diagonals. For every vertex the number of sides issuing from it is even, except for the vertices
A
i
1
,
A
i
2
,
…
,
A
i
k
A_{i_1},A_{i_2},\dots,A_{i_k}
A
i
1
,
A
i
2
,
…
,
A
i
k
, where
1
≤
i
1
<
⋯
<
i
k
≤
n
1\leq i_1<\dots<i_k\leq n
1
≤
i
1
<
⋯
<
i
k
≤
n
. Prove that
k
k
k
is even and
n
≡
i
1
−
i
2
+
⋯
+
i
k
−
1
−
i
k
(
m
o
d
3
)
n\equiv i_1-i_2+\dots+i_{k-1}-i_k\pmod3
n
≡
i
1
−
i
2
+
⋯
+
i
k
−
1
−
i
k
(
mod
3
)
if
k
>
0
k>0
k
>
0
and n\equiv0\pmod3\mbox{ for }k=0. Note that this leads to generalization of one recent Tournament of towns problem about triangulating of square.
kings tour and dominoes
An
8
×
8
8\times8
8
×
8
chessboard is completely tiled by
2
×
1
2\times1
2
×
1
dominoes. Prove that there exist a king's tour of that chessboard such that every cell of the board is visited exactly once and such that king goes domino by domino, i.e. if king moves to the first cell of a domino, it must move to another cell in the next move. (King doesn't have to come back to the initial cell. King is an usual chess piece.)
1
3
Hide problems
very easy
Prove that the cube of any positive integer greater than 1 can be represented as a difference of the squares of two positive integers.
equation
Find all real solutions of the equation
x
−
1
3
+
3
x
−
1
3
=
x
+
1
3
.
\sqrt[3]{x-1}+\sqrt[3]{3x-1}=\sqrt[3]{x+1}.
3
x
−
1
+
3
3
x
−
1
=
3
x
+
1
.
polynomial and sequence
Define the sequence
(
a
n
)
n
≥
1
(a_n)_{n\geq 1}
(
a
n
)
n
≥
1
by
a
1
=
1
a_1=1
a
1
=
1
,
a
2
=
p
a_2=p
a
2
=
p
and
a
n
+
1
=
p
a
n
−
a
n
−
1
for all
n
>
1.
a_{n+1}=pa_n-a_{n-1} \textrm { for all } n>1.
a
n
+
1
=
p
a
n
−
a
n
−
1
for all
n
>
1.
Prove that for
n
>
1
n>1
n
>
1
the polynomial
x
n
−
a
n
x
+
a
n
−
1
x^n-a_nx+a_{n-1}
x
n
−
a
n
x
+
a
n
−
1
is divisible by
x
2
−
p
x
+
1
x^2-px+1
x
2
−
p
x
+
1
. Using this result, solve the equation
x
4
−
56
x
+
15
=
0.
x^4-56x+15=0.
x
4
−
56
x
+
15
=
0.
2
4
Show problems
3
4
Show problems