MathDB
Problems
Contests
National and Regional Contests
Moldova Contests
Moldova Team Selection Test
1993 Moldova Team Selection Test
1993 Moldova Team Selection Test
Part of
Moldova Team Selection Test
Subcontests
(7)
9
1
Hide problems
Determine the greatest integer $m$ for which $2102$ is $m$-additive
Positive integer
q
q{}
q
is
m
−
a
d
d
i
t
i
v
e
m-additive
m
−
a
dd
i
t
i
v
e
(
m
∈
N
,
m
≥
2
)
(m\in\mathbb{N}, m\geq2)
(
m
∈
N
,
m
≥
2
)
if there exist pairwise distinct positive integers
a
1
,
a
2
,
…
,
a
m
a_1,a_2,\ldots,a_m
a
1
,
a
2
,
…
,
a
m
such that
q
=
a
1
+
a
2
+
…
+
a
m
q=a_1+a_2+\ldots+a_m
q
=
a
1
+
a
2
+
…
+
a
m
and
a
i
∣
a
i
+
1
a_i | a_{i+1}
a
i
∣
a
i
+
1
for
i
=
1
,
2
,
…
,
m
−
1
i=1,2,\ldots,m-1
i
=
1
,
2
,
…
,
m
−
1
. a) Prove that
1993
1993
1993
is
8
8
8
-additive, but
9
9
9
-additive. b) Determine the greatest integer
m
m
m
for which
2102
2102
2102
is
m
m
m
-additive.
8
1
Hide problems
rove that the quadrilateral $O_1O_2O_3O_4$ is a parallelogram.
Inside the parallelogram
A
B
C
D
ABCD
A
BC
D
points
M
,
N
,
K
M, N, K
M
,
N
,
K
and
L
L{}
L
are on sides
A
B
,
B
C
,
C
D
AB, BC, CD{}
A
B
,
BC
,
C
D
and
D
A
DA
D
A
, respectively. Let
O
1
,
O
2
,
O
3
O_1, O_2, O_3
O
1
,
O
2
,
O
3
and
O
4
O_4
O
4
be the circumcenters of triangles repesctively
M
B
N
,
N
C
K
,
K
D
L
MBN, NCK, KDL
MBN
,
NC
K
,
KD
L
and
L
A
M
LAM{}
L
A
M
. Prove that the quadrilateral
O
1
O
2
O
3
O
4
O_1O_2O_3O_4
O
1
O
2
O
3
O
4
is a parallelogram.
5
1
Hide problems
Inside the acute triangle $ABC$ the point $P{}$ in on height $AA_1{}$.
Inside the acute triangle
A
B
C
ABC
A
BC
the point
P
P{}
P
in on height
A
A
1
AA_1{}
A
A
1
. Lines
B
P
BP{}
BP
and
C
P
CP{}
CP
intersect the sides
A
C
AC{}
A
C
and
A
B
AB{}
A
B
, respectively, in points
B
1
B_1{}
B
1
and
C
1
C_1{}
C
1
. Prove that: a)
A
A
1
AA_1{}
A
A
1
is the bisector of the angle
B
1
A
1
C
1
;
B_1A_1C_1;
B
1
A
1
C
1
;
b) if the lines
B
C
BC
BC
and
B
1
C
1
B_1C_1
B
1
C
1
are concurrent, then the position of theri intersection does not depend on
P
.
P.
P
.
4
1
Hide problems
Solve in positive integers the following equation
Solve in positive integers the following equation
[
1
]
+
[
2
]
+
[
3
]
+
…
+
[
x
2
−
2
]
+
[
x
2
−
1
]
=
125
,
\left [\sqrt{1}\right]+\left [\sqrt{2}\right]+\left [\sqrt{3}\right]+\ldots+\left [\sqrt{x^2-2}\right]+\left [\sqrt{x^2-1}\right]=125,
[
1
]
+
[
2
]
+
[
3
]
+
…
+
[
x
2
−
2
]
+
[
x
2
−
1
]
=
125
,
where
[
a
]
[a]
[
a
]
is the integer part of the real number
a
a
a
.
3
1
Hide problems
$g(x)=C\cdot10^{-|x-d|}$
Let
f
:
R
→
R
f:\mathbb{R}\rightarrow\mathbb{R}
f
:
R
→
R
be a function defined as the maximum of a finite number of functions
g
:
R
→
R
g:\mathbb{R}\rightarrow\mathbb{R}
g
:
R
→
R
of the form
g
(
x
)
=
C
⋅
1
0
−
∣
x
−
d
∣
g(x)=C\cdot10^{-|x-d|}
g
(
x
)
=
C
⋅
1
0
−
∣
x
−
d
∣
(with different values of parameters
d
d{}
d
and
C
>
0
C>0
C
>
0
). For real numbers
a
<
b
a<b
a
<
b
we have
f
(
a
)
=
f
(
b
)
f(a)=f(b)
f
(
a
)
=
f
(
b
)
. Prove that on the segment
[
a
;
b
]
[a;b]
[
a
;
b
]
the sum of legnths of segments on which
f
f
f
is increasing is equal to the sum of legnths of segments on which
f
f
f
is decreasing.
6
1
Hide problems
Combo seems Easy but I cannot Get
The numbers
1
,
2
,
.
.
.
,
2
n
−
1
,
2
n
1,2,...,2n-1,2n
1
,
2
,
...
,
2
n
−
1
,
2
n
are divided into two disjoint sets,
a
1
<
a
2
<
.
.
.
<
a
n
a_1 < a_2 < ... < a_n
a
1
<
a
2
<
...
<
a
n
and
b
1
>
b
2
>
.
.
.
>
b
n
b_1 > b_2 > ... > b_n
b
1
>
b
2
>
...
>
b
n
. Prove that
∣
a
1
−
b
1
∣
+
∣
a
2
−
b
2
∣
+
.
.
.
+
∣
a
n
−
b
n
∣
=
n
2
.
|a_1 - b_1| + |a_2 - b_2| + ... + |a_n - b_n| = n^2.
∣
a
1
−
b
1
∣
+
∣
a
2
−
b
2
∣
+
...
+
∣
a
n
−
b
n
∣
=
n
2
.
7
1
Hide problems
inequality
If
x
1
+
x
2
+
⋯
+
x
n
=
∑
i
=
1
n
x
i
=
1
2
x_1 + x_2 + \cdots + x_n = \sum_{i=1}^{n} x_i = \frac{1}{2}
x
1
+
x
2
+
⋯
+
x
n
=
∑
i
=
1
n
x
i
=
2
1
and
x
i
>
0
x_i > 0
x
i
>
0
; then prove that:
1
−
x
1
1
+
x
1
⋅
1
−
x
2
1
+
x
2
⋯
1
−
x
n
1
+
x
n
=
∏
i
=
1
n
1
−
x
i
1
+
x
i
≥
1
3
\frac{1-x_1}{1+x_1} \cdot \frac{1-x_2}{1+x_2} \cdots \frac{1-x_n}{1+x_n} = \prod_{i=1}^{n} \frac{1-x_i}{1+x_i} \geq \frac{1}{3}
1
+
x
1
1
−
x
1
⋅
1
+
x
2
1
−
x
2
⋯
1
+
x
n
1
−
x
n
=
∏
i
=
1
n
1
+
x
i
1
−
x
i
≥
3
1