MathDB

1

divided circle

13

segments, numbered consecutively from

1

to

13

. Five fleas called

A,B,C,D

and

E

are sitting in the segments

1,2,3,4

and

5

. A flea is allowed to jump to an empty segment five positions away in either direction around the circle. Only one flea jumps at the same time, and two fleas cannot be in the same segment. After some jumps, the fleas are back in the segments

1,2,3,4,5

, but possibly in some other order than they started. Which orders are possible ?

13

1

European Union

The

25

member states of the European Union set up a committee with the following rules: 1) the committee should meet daily; 2) at each meeting, at least one member should be represented; 3) at any two different meetings, a different set of member states should be represented; 4) at

n^{th}

meeting, for every

k<n

, the set of states represented should include at least one state that was represented at the

k^{th}

meeting.
For how many days can the committee have its meetings?

11

1

Table m x n

Given a table

m\times n

, in each cell of which a number

+1

or

-1

is written. It is known that initially exactly one

-1

is in the table, all the other numbers being

+1

. During a move, it is allowed to chose any cell containing

-1

, replace this

-1

by

0

, and simultaneously multiply all the numbers in the neighbouring cells by

-1

(we say that two cells are neighbouring if they have a common side). Find all

(m,n)

for which using such moves one can obtain the table containing zeros only, regardless of the cell in which the initial

-1

stands.

12

1

interchanging numbers

There are

2n

different numbers in a row. By one move we can interchange any two numbers or interchange any

3

numbers cyclically (choose

a,b,c

and place

a

instead of

b

,

b

instead of

c

,

c

instead of

a

). What is the minimal number of moves that is always sufficient to arrange the numbers in increasing order ?

9

1

set of natural numbers

A set

S

of

n-1

natural numbers is given (

n\ge 3

). There exist at least at least two elements in this set whose difference is not divisible by

n

. Prove that it is possible to choose a non-empty subset of

S

so that the sum of its elements is divisible by

n

.

5

1

Function

Determine the range of the following function defined for integer

k

,

f(k)=(k)_3+(2k)_5+(3k)_7-6k

where

(k)_{2n+1}

denotes the multiple of

2n+1

closest to

k

1

A sequence

Given a sequence

a_1,a_2,\ldots

of non-negative real numbers satisfying the conditions:
1.

a_n + a_{2n} \geq 3n

; 2.

a_{n+1}+n \leq 2\sqrt{a_n \left(n+1\right)}

for all

n\in\mathbb N

(where

\mathbb N=\left\{1,2,3,...\right\}

).
(1) Prove that the inequality

a_n \geq n

holds for every

n \in \mathbb N

. (2) Give an example of such a sequence.

7

1

More funny number theory

Find all sets

X

consisting of at least two positive integers such that for every two elements

m,n\in X

, where

n>m

, there exists an element

k\in X

such that

n=mk^2

.

6

1

Natural numbers on the faces of a cube

A positive integer is written on each of the six faces of a cube. For each vertex of the cube we compute the product of the numbers on the three adjacent faces. The sum of these products is

1001

. What is the sum of the six numbers on the faces?

8

1

Beautiful polynomial number theory

Let

f\left(x\right)

be a non-constant polynomial with integer coefficients, and let

u

be an arbitrary positive integer. Prove that there is an integer

n

such that

f\left(n\right)

has at least

u

distinct prime factors and

f\left(n\right) \neq 0

.

14

1

a pile is a set of four or more nuts

We say that a pile is a set of four or more nuts. Two persons play the following game. They start with one pile of

n \geq 4

nuts. During a move a player takes one of the piles that they have and split it into two nonempty sets (these sets are not necessarily piles, they can contain arbitrary number of nuts). If the player cannot move, he loses. For which values of

n

does the first player have a winning strategy?

4

1

Arithmetic means of subsequences bounded by arithmetic mean

Let

x_1

,

x_2

, ...,

x_n

be real numbers with arithmetic mean

X

. Prove that there is a positive integer

K

such that for any integer

i

satisfying

0\leq i<K

, we have

\frac{1}{K-i}\sum_{j=i+1}^{K} x_j \leq X

. (In other words, prove that there is a positive integer

K

such that the arithmetic mean of each of the lists

\left\{x_1,x_2,...,x_K\right\}

,

\left\{x_2,x_3,...,x_K\right\}

,

\left\{x_3,...,x_K\right\}

, ...,

\left\{x_{K-1},x_K\right\}

,

\left\{x_K\right\}

is not greater than

X

.)

3

1

A nice joke

Let

p, q, r

be positive real numbers and

n

a natural number. Show that if

pqr = 1

, then

\frac{1}{p^n+q^n+1} + \frac{1}{q^n+r^n+1} + \frac{1}{r^n+p^n+1} \leq 1.

17

1

Anything but geometry

Consider a rectangle with sidelengths 3 and 4, pick an arbitrary inner point on each side of this rectangle. Let

x, y, z

and

u

denote the side lengths of the quadrilateral spanned by these four points. Prove that

25 \leq x^2+y^2+z^2+u^2 \leq 50

.

18

1

And this is called geometry?!

A ray emanating from the vertex

A

of the triangle

ABC

intersects the side

BC

at

X

and the circumcircle of triangle

ABC

at

Y

. Prove that

\frac{1}{AX}+\frac{1}{XY}\geq \frac{4}{BC}

.

16

1

Diameter, tangent and chord

Through a point

P

exterior to a given circle pass a secant and a tangent to the circle. The secant intersects the circle at

A

and

B

, and the tangent touches the circle at

C

on the same side of the diameter through

P

as the points

A

and

B

. The projection of the point

C

on the diameter is called

Q

. Prove that the line

QC

bisects the angle

\angle AQB

.

19

1

Symmedian and circumcircles

Let

D

be the midpoint of the side

BC

of a triangle

ABC

. Let

M

be a point on the side

BC

such that

\angle BAM = \angle DAC

. Further, let

L

be the second intersection point of the circumcircle of the triangle

CAM

with the side

AB

, and let

K

be the second intersection point of the circumcircle of the triangle

BAM

with the side

AC

. Prove that

KL \parallel BC

.

20

1

XY/YZ is independent of the position of X.

Three fixed circles pass through the points

A

and

B

. Let

X

be a variable point on the first circle different from

A

and

B

. The line

AX

intersects the other two circles at

Y

and

Z

(with

Y

between

X

and

Z

). Show that the ratio

\frac{XY}{YZ}

is independent of the position of

X

.

2

1

P(1/x) * P(x) >= 1 for x = 1 ==> also for every x > 0

Let

P(x)

be a polynomial with a non-negative coefficients. Prove that if the inequality

P\left(\frac {1}{x}\right)P(x)\geq 1

holds for x \equal{} 1, then this inequality holds for each positive

x

.

10

1

The most beautiful problem

Is there an infinite sequence of prime numbers

p_1

,

p_2

,

\ldots

,

p_n

,

p_{n+1}

,

\ldots

such that

|p_{n+1}-2p_n|=1

for each

n \in \mathbb{N}

?

2004 Baltic Way

Subcontests

divided circle

European Union

Table m x n

interchanging numbers

set of natural numbers

Function

A sequence

More funny number theory

Natural numbers on the faces of a cube

Beautiful polynomial number theory

a pile is a set of four or more nuts

Arithmetic means of subsequences bounded by arithmetic mean

A nice joke

Anything but geometry

And this is called geometry?!

Diameter, tangent and chord

Symmedian and circumcircles

XY/YZ is independent of the position of X.

P(1/x) * P(x) >= 1 for x = 1 ==> also for every x > 0

The most beautiful problem

2004 Baltic Way

Subcontests

divided circle

European Union

Table m x n

interchanging numbers

set of natural numbers

Function

A sequence

More funny number theory

Natural numbers on the faces of a cube

Beautiful polynomial number theory

a pile is a set of four or more nuts

Arithmetic means of subsequences bounded by arithmetic mean

A nice joke

Anything but geometry

And this is called geometry?!

Diameter, tangent and chord

Symmedian and circumcircles

XY/YZ is independent of the position of X.

P(1/x) * P(x) &gt;= 1 for x = 1 ==&gt; also for every x &gt; 0

The most beautiful problem

P(1/x) * P(x) >= 1 for x = 1 ==> also for every x > 0