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Problems
Contests
National and Regional Contests
Moldova Contests
Moldova Team Selection Test
2004 Moldova Team Selection Test
2004 Moldova Team Selection Test
Part of
Moldova Team Selection Test
Subcontests
(11)
12
1
Hide problems
Show that $a_{2m}=10a_{2m-1}
Let
a
k
a_k
a
k
be the number of nonnegative integers
n
n
n
with the properties: a)
n
∈
[
0
,
1
0
k
)
n\in[0, 10^k)
n
∈
[
0
,
1
0
k
)
has exactly
k
k
k
digits, such that he zeroes on the first positions of
n
n
n
are included in the decimal writting. b) the digits of
n
n
n
can be permutated such that the new number is divisible by
11.
11.
11.
Show that
a
2
m
=
10
a
2
m
−
1
a_{2m}=10a_{2m-1}
a
2
m
=
10
a
2
m
−
1
for every
m
∈
N
.
m\in\mathbb{N}.
m
∈
N
.
8
1
Hide problems
An integer $n$ is called good if $|n|$ is not a square of an integer.
An integer
n
n
n
is called good if
∣
n
∣
|n|
∣
n
∣
is not a square of an integer. Find all integers
m
m
m
with the following property:
m
m
m
can be represented in infinite ways as a sum of three disctinct good numbers, the product of which is the square of an odd integer.
5
1
Hide problems
f(x_1,...,x_n)=\frac{1}{x_1}+\frac{1}{2x_2}+\ldots+\frac{1}{(n-1)x_{n-1}}+\frac{
Let
n
∈
N
n\in\mathbb{N}
n
∈
N
, the set
A
=
{
(
x
1
,
x
2
.
.
.
,
x
n
)
∣
x
i
∈
R
+
,
i
=
1
,
2
,
.
.
.
,
n
}
A=\{(x_1,x_2...,x_n)|x_i\in\mathbb{R}_{+}, i=1,2,...,n\}
A
=
{(
x
1
,
x
2
...
,
x
n
)
∣
x
i
∈
R
+
,
i
=
1
,
2
,
...
,
n
}
and the function
f
:
A
→
R
,
f
(
x
1
,
.
.
.
,
x
n
)
=
1
x
1
+
1
2
x
2
+
…
+
1
(
n
−
1
)
x
n
−
1
+
1
n
x
n
.
f:A\rightarrow\mathbb{R}, f(x_1,...,x_n)=\frac{1}{x_1}+\frac{1}{2x_2}+\ldots+\frac{1}{(n-1)x_{n-1}}+\frac{1}{nx_n}.
f
:
A
→
R
,
f
(
x
1
,
...
,
x
n
)
=
x
1
1
+
2
x
2
1
+
…
+
(
n
−
1
)
x
n
−
1
1
+
n
x
n
1
.
Prove that
f
(
(
n
1
)
,
(
n
2
)
,
.
.
.
,
(
n
n
−
1
)
,
(
n
n
)
)
=
f
(
2
n
−
1
,
2
n
−
2
,
.
.
.
,
2
,
1
)
.
f(\textstyle\binom{n}{1},\binom{n}{2},...,\binom{n}{n-1},\binom{n}{n})=f(2^{n-1},2^{n-2},...,2,1).
f
(
(
1
n
)
,
(
2
n
)
,
...
,
(
n
−
1
n
)
,
(
n
n
)
)
=
f
(
2
n
−
1
,
2
n
−
2
,
...
,
2
,
1
)
.
3
1
Hide problems
The cricles $\Gamma_1$ and $\Gamma_2$ intersect in $M$ and $N$.
The cricles
Γ
1
\Gamma_1
Γ
1
and
Γ
2
\Gamma_2
Γ
2
intersect in
M
M
M
and
N
.
N.
N
.
A line that goes through
M
M
M
intersects the cricles
Γ
1
\Gamma_1
Γ
1
and
Γ
2
\Gamma_2
Γ
2
in
A
A
A
and
B
B
B
, such that
M
∈
(
A
B
)
M\in(AB)
M
∈
(
A
B
)
. The bisector of angle
A
M
N
AMN
A
MN
intersects the circle
Γ
1
\Gamma_1
Γ
1
in
D
,
D,
D
,
and the bisector of angle
B
M
N
BMN
BMN
intersects the circle
Γ
2
\Gamma_2
Γ
2
in
C
.
C.
C
.
Prove that the circle with diameter
C
D
CD
C
D
splits the segment
A
B
AB
A
B
in half.
2
1
Hide problems
\frac{1}{\sqrt{r_A^2-r_Ar_B+r_B^2}}+\frac{1}{\sqrt{r_B^2-r_Br_C+r_C^2}}
In the tetrahedron
A
B
C
D
ABCD
A
BC
D
the radius of its inscribed sphere is
r
r
r
and the radiuses of the exinscribed spheres (each tangent with a face of the tetrahedron and with the planes of the other faces) are
r
A
,
r
B
,
r
C
,
r
D
.
r_A, r_B, r_C, r_D.
r
A
,
r
B
,
r
C
,
r
D
.
Prove the inequality
1
r
A
2
−
r
A
r
B
+
r
B
2
+
1
r
B
2
−
r
B
r
C
+
r
C
2
+
1
r
C
2
−
r
C
r
D
+
r
D
2
+
1
r
D
2
−
r
D
r
A
+
r
A
2
≤
2
r
.
\frac{1}{\sqrt{r_A^2-r_Ar_B+r_B^2}}+\frac{1}{\sqrt{r_B^2-r_Br_C+r_C^2}}+\frac{1}{\sqrt{r_C^2-r_Cr_D+r_D^2}}+\frac{1}{\sqrt{r_D^2-r_Dr_A+r_A^2}}\leq\frac{2}{r}.
r
A
2
−
r
A
r
B
+
r
B
2
1
+
r
B
2
−
r
B
r
C
+
r
C
2
1
+
r
C
2
−
r
C
r
D
+
r
D
2
1
+
r
D
2
−
r
D
r
A
+
r
A
2
1
≤
r
2
.
4
1
Hide problems
Combinatorics involving functions.
Let
n
n
n
be an integer bigger than
0
0
0
. Let
A
=
(
a
1
,
a
2
,
.
.
.
,
a
n
)
\mathbb{A}= ( a_1,a_2,...,a_n )
A
=
(
a
1
,
a
2
,
...
,
a
n
)
be a set of real numbers. Find the number of functions
f
:
A
→
A
f:A \rightarrow A
f
:
A
→
A
such that
f
(
f
(
x
)
)
−
f
(
f
(
y
)
)
≥
x
−
y
f(f(x))-f(f(y)) \ge x-y
f
(
f
(
x
))
−
f
(
f
(
y
))
≥
x
−
y
for any
x
,
y
∈
A
x,y \in \mathbb{A}
x
,
y
∈
A
, with
x
>
y
x>y
x
>
y
.
6
1
Hide problems
Easy function (I guess)
Find all functions
f
:
R
→
R
f:\mathbb R \to \mathbb R
f
:
R
→
R
Such that for all real
x
,
y
x,y
x
,
y
:
(
x
2
+
x
y
+
y
2
)
(
f
(
x
)
−
f
(
y
)
)
=
f
(
x
3
)
−
f
(
y
3
)
(x^2+xy+y^2)(f(x)-f(y))=f(x^3)-f(y^3)
(
x
2
+
x
y
+
y
2
)
(
f
(
x
)
−
f
(
y
))
=
f
(
x
3
)
−
f
(
y
3
)
9
1
Hide problems
2004 Republic of Moldova , Problem 9
Let
a
,
b
a,b
a
,
b
and
c
c
c
be positive real numbers . Prove that
∣
4
(
b
3
−
c
3
)
b
+
c
+
4
(
c
3
−
a
3
)
c
+
a
+
4
(
a
3
−
b
3
)
a
+
b
∣
≤
(
b
−
c
)
2
+
(
c
−
a
)
2
+
(
a
−
b
)
2
.
\left | \frac{4(b^3-c^3)}{b+c}+ \frac{4(c^3-a^3)}{c+a}+ \frac{4(a^3-b^3)}{a+b} \right |\leq (b-c)^2+(c-a)^2+(a-b)^2.
b
+
c
4
(
b
3
−
c
3
)
+
c
+
a
4
(
c
3
−
a
3
)
+
a
+
b
4
(
a
3
−
b
3
)
≤
(
b
−
c
)
2
+
(
c
−
a
)
2
+
(
a
−
b
)
2
.
10
1
Hide problems
Polynomial Equation
Determine all polynomials
P
(
x
)
P(x)
P
(
x
)
with real coeffcients such that
(
x
3
+
3
x
2
+
3
x
+
2
)
P
(
x
−
1
)
=
(
x
3
−
3
x
2
+
3
x
−
2
)
P
(
x
)
(x^3+3x^2+3x+2)P(x-1)=(x^3-3x^2+3x-2)P(x)
(
x
3
+
3
x
2
+
3
x
+
2
)
P
(
x
−
1
)
=
(
x
3
−
3
x
2
+
3
x
−
2
)
P
(
x
)
.
1
1
Hide problems
Number theory Marathon
Hello Everyone, i'm trying to make a strong marathon for number theory .. which will be in Pre-Olympiad levelPlease if you write any problem don't forget to indicate its number and if you write a solution please indicate for what problem also to prevent the confusion that happens in some marathons. it will be preferred to write the source of the problem. please , show detailed solutions , and please post some challenging Pre-Olympiad problems.. remember ,, different approaches are most welcome :)now .. let's get our hands dirty :lol: : let
f
(
n
)
f(n)
f
(
n
)
denote the sum of the digits of
n
n
n
. Now let N \equal{} {4444^{4444}}. Find
f
(
f
(
f
(
N
)
)
)
f\left( {f\left( {f\left( N \right)} \right)} \right)
f
(
f
(
f
(
N
)
)
)
.
7
1
Hide problems
Geometrical inequality, similar to treegoner's one
Let
A
B
C
ABC
A
BC
be a triangle, let
O
O
O
be its circumcenter, and let
H
H
H
be its orthocenter. Let
P
P
P
be a point on the segment
O
H
OH
O
H
. Prove that
6
r
≤
P
A
+
P
B
+
P
C
≤
3
R
6r\leq PA+PB+PC\leq 3R
6
r
≤
P
A
+
PB
+
PC
≤
3
R
, where
r
r
r
is the inradius and
R
R
R
the circumradius of triangle
A
B
C
ABC
A
BC
. Moderator edit: This is true only if the point
P
P
P
lies inside the triangle
A
B
C
ABC
A
BC
. (Of course, this is always fulfilled if triangle
A
B
C
ABC
A
BC
is acute-angled, since in this case the segment
O
H
OH
O
H
completely lies inside the triangle
A
B
C
ABC
A
BC
; but if triangle
A
B
C
ABC
A
BC
is obtuse-angled, then the condition about
P
P
P
lying inside the triangle
A
B
C
ABC
A
BC
is really necessary.)