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Contests
National and Regional Contests
Romania Contests
Romania Team Selection Test
2003 Romania Team Selection Test
2003 Romania Team Selection Test
Part of
Romania Team Selection Test
Subcontests
(17)
18
1
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For n the sum of digits =d(n), then x+d(x)=m has k solutions
For every positive integer
n
n
n
we denote by
d
(
n
)
d(n)
d
(
n
)
the sum of its digits in the decimal representation. Prove that for each positive integer
k
k
k
there exists a positive integer
m
m
m
such that the equation
x
+
d
(
x
)
=
m
x+d(x)=m
x
+
d
(
x
)
=
m
has exactly
k
k
k
solutions in the set of positive integers.
16
1
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Given a convex hexagon and 6 triangle areas find its area
Let
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
be a convex hexagon and denote by
A
′
,
B
′
,
C
′
,
D
′
,
E
′
,
F
′
A',B',C',D',E',F'
A
′
,
B
′
,
C
′
,
D
′
,
E
′
,
F
′
the middle points of the sides
A
B
AB
A
B
,
B
C
BC
BC
,
C
D
CD
C
D
,
D
E
DE
D
E
,
E
F
EF
EF
and
F
A
FA
F
A
respectively. Given are the areas of the triangles
A
B
C
′
ABC'
A
B
C
′
,
B
C
D
′
BCD'
BC
D
′
,
C
D
E
′
CDE'
C
D
E
′
,
D
E
F
′
DEF'
D
E
F
′
,
E
F
A
′
EFA'
EF
A
′
and
F
A
B
′
FAB'
F
A
B
′
. Find the area of the hexagon. Kvant Magazine
14
1
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Rhombus ABCD of side 1 - find the measures of the angles
Given is a rhombus
A
B
C
D
ABCD
A
BC
D
of side 1. On the sides
B
C
BC
BC
and
C
D
CD
C
D
we are given the points
M
M
M
and
N
N
N
respectively, such that
M
C
+
C
N
+
M
N
=
2
MC+CN+MN=2
MC
+
CN
+
MN
=
2
and
2
∠
M
A
N
=
∠
B
A
D
2\angle MAN = \angle BAD
2∠
M
A
N
=
∠
B
A
D
. Find the measures of the angles of the rhombus. Cristinel Mortici
13
1
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Senators are arranged in 10 parties and committees
A parliament has
n
n
n
senators. The senators form 10 parties and 10 committees, such that any senator belongs to exactly one party and one committee. Find the least possible
n
n
n
for which it is possible to label the parties and the committees with numbers from 1 to 10, such that there are at least 11 senators for which the numbers of the corresponding party and committee are equal.
9
1
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Array nxn with positive integers that add up to n^3
Let
n
≥
3
n\geq 3
n
≥
3
be a positive integer. Inside a
n
×
n
n\times n
n
×
n
array there are placed
n
2
n^2
n
2
positive numbers with sum
n
3
n^3
n
3
. Prove that we can find a square
2
×
2
2\times 2
2
×
2
of 4 elements of the array, having the sides parallel with the sides of the array, and for which the sum of the elements in the square is greater than
3
n
3n
3
n
. Radu Gologan
8
1
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Two externally tangent circles and two tangent paralel lines
Two circles
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
with radii
r
1
r_1
r
1
and
r
2
r_2
r
2
,
r
2
>
r
1
r_2>r_1
r
2
>
r
1
, are externally tangent. The line
t
1
t_1
t
1
is tangent to the circles
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
at points
A
A
A
and
D
D
D
respectively. The parallel line
t
2
t_2
t
2
to the line
t
1
t_1
t
1
is tangent to the circle
ω
1
\omega_1
ω
1
and intersects the circle
ω
2
\omega_2
ω
2
at points
E
E
E
and
F
F
F
. The line
t
3
t_3
t
3
passing through
D
D
D
intersects the line
t
2
t_2
t
2
and the circle
ω
2
\omega_2
ω
2
in
B
B
B
and
C
C
C
respectively, both different of
E
E
E
and
F
F
F
respectively. Prove that the circumcircle of the triangle
A
B
C
ABC
A
BC
is tangent to the line
t
1
t_1
t
1
. Dinu Serbanescu
7
1
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Find a,b,m,n such that x^n+ax+b divides x^m+ax+b
Find all integers
a
,
b
,
m
,
n
a,b,m,n
a
,
b
,
m
,
n
, with
m
>
n
>
1
m>n>1
m
>
n
>
1
, for which the polynomial
f
(
X
)
=
X
n
+
a
X
+
b
f(X)=X^n+aX+b
f
(
X
)
=
X
n
+
a
X
+
b
divides the polynomial
g
(
X
)
=
X
m
+
a
X
+
b
g(X)=X^m+aX+b
g
(
X
)
=
X
m
+
a
X
+
b
. Laurentiu Panaitopol
6
1
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2n students at a "fair" math competition
At a math contest there are
2
n
2n
2
n
students participating. Each of them submits a problem to the jury, which thereafter gives each students one of the
2
n
2n
2
n
problems submitted. One says that the contest is fair is there are
n
n
n
participants which receive their problems from the other
n
n
n
participants. Prove that the number of distributions of the problems in order to obtain a fair contest is a perfect square.
5
1
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If f is irreducible and |f(0)| not a square then f(x^2) irr.
Let
f
∈
Z
[
X
]
f\in\mathbb{Z}[X]
f
∈
Z
[
X
]
be an irreducible polynomial over the ring of integer polynomials, such that
∣
f
(
0
)
∣
|f(0)|
∣
f
(
0
)
∣
is not a perfect square. Prove that if the leading coefficient of
f
f
f
is 1 (the coefficient of the term having the highest degree in
f
f
f
) then
f
(
X
2
)
f(X^2)
f
(
X
2
)
is also irreducible in the ring of integer polynomials. Mihai Piticari
4
1
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Geometric progression among the numbers n\sqrt 2003
Prove that among the elements of the sequence
{
⌊
n
2003
⌋
}
n
≥
1
\left\{ \left\lfloor n\sqrt{2003} \right\rfloor \right\}_{n\geq 1}
{
⌊
n
2003
⌋
}
n
≥
1
one can find a geometric progression having any number of terms, and having the ratio bigger than
k
k
k
, where
k
k
k
can be any positive integer. Radu Gologan
3
1
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Just for the record ...
Let
n
,
k
n,k
n
,
k
be positive integers such that
n
k
>
(
k
+
1
)
!
n^k>(k+1)!
n
k
>
(
k
+
1
)!
and consider the set M=\{(x_1,x_2,\ldots,x_n)\dvd x_i\in\{1,2,\ldots,n\},\ i=\overline{1,k}\}. Prove that if
A
⊂
M
A\subset M
A
⊂
M
has
(
k
+
1
)
!
+
1
(k+1)!+1
(
k
+
1
)!
+
1
elements, then there are two elements
{
α
,
β
}
⊂
A
\{\alpha,\beta\}\subset A
{
α
,
β
}
⊂
A
,
α
=
(
α
1
,
α
2
,
…
,
α
n
)
\alpha=(\alpha_1,\alpha_2,\ldots,\alpha_n)
α
=
(
α
1
,
α
2
,
…
,
α
n
)
,
β
=
(
β
1
,
β
2
,
…
,
β
n
)
\beta=(\beta_1,\beta_2,\ldots,\beta_n)
β
=
(
β
1
,
β
2
,
…
,
β
n
)
such that
(
k
+
1
)
!
∣
(
β
1
−
α
1
)
(
β
2
−
α
2
)
⋯
(
β
k
−
α
k
)
.
(k+1)! \left| (\beta_1-\alpha_1)(\beta_2-\alpha_2)\cdots (\beta_k-\alpha_k) \right. .
(
k
+
1
)!
∣
(
β
1
−
α
1
)
(
β
2
−
α
2
)
⋯
(
β
k
−
α
k
)
.
12
1
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Simple words
A word is a sequence of n letters of the alphabet {a, b, c, d}. A word is said to be complicated if it contains two consecutive groups of identic letters. The words caab, baba and cababdc, for example, are complicated words, while bacba and dcbdc are not. A word that is not complicated is a simple word. Prove that the numbers of simple words with n letters is greater than
2
n
2^n
2
n
, if n is a positive integer.
15
1
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invisible lattice points
In a plane we choose a cartesian system of coordinates. A point
A
(
x
,
y
)
A(x,y)
A
(
x
,
y
)
in the plane is called an integer point if and only if both
x
x
x
and
y
y
y
are integers. An integer point
A
A
A
is called invisible if on the segment
(
O
A
)
(OA)
(
O
A
)
there is at least one integer point. Prove that for each positive integer
n
n
n
there exists a square of side
n
n
n
in which all the interior integer points are invisible.
2
1
Hide problems
I think is little hard
Let
A
B
C
ABC
A
BC
be a triangle with
∠
B
A
C
=
6
0
∘
\angle BAC=60^\circ
∠
B
A
C
=
6
0
∘
. Consider a point
P
P
P
inside the triangle having
P
A
=
1
PA=1
P
A
=
1
,
P
B
=
2
PB=2
PB
=
2
and
P
C
=
3
PC=3
PC
=
3
. Find the maximum possible area of the triangle
A
B
C
ABC
A
BC
.
10
1
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a nice problem
Let
P
\mathcal{P}
P
be the set of all primes, and let
M
M
M
be a subset of
P
\mathcal{P}
P
, having at least three elements, and such that for any proper subset
A
A
A
of
M
M
M
all of the prime factors of the number
−
1
+
∏
p
∈
A
p
-1+\prod_{p\in A}p
−
1
+
∏
p
∈
A
p
are found in
M
M
M
. Prove that
M
=
P
M= \mathcal{P}
M
=
P
.Valentin Vornicu
17
1
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straight permutation
A permutation
σ
:
{
1
,
2
,
…
,
n
}
→
{
1
,
2
,
…
,
n
}
\sigma: \{1,2,\ldots,n\}\to\{1,2,\ldots,n\}
σ
:
{
1
,
2
,
…
,
n
}
→
{
1
,
2
,
…
,
n
}
is called straight if and only if for each integer
k
k
k
,
1
≤
k
≤
n
−
1
1\leq k\leq n-1
1
≤
k
≤
n
−
1
the following inequality is fulfilled
∣
σ
(
k
)
−
σ
(
k
+
1
)
∣
≤
2.
|\sigma(k)-\sigma(k+1)|\leq 2.
∣
σ
(
k
)
−
σ
(
k
+
1
)
∣
≤
2.
Find the smallest positive integer
n
n
n
for which there exist at least 2003 straight permutations. Valentin Vornicu
11
1
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Square of side 6 has 4 points at dist. 5 from each other in
In a square of side 6 the points
A
,
B
,
C
,
D
A,B,C,D
A
,
B
,
C
,
D
are given such that the distance between any two of the four points is at least 5. Prove that
A
,
B
,
C
,
D
A,B,C,D
A
,
B
,
C
,
D
form a convex quadrilateral and its area is greater than 21. Laurentiu Panaitopol