MathDB

2

Three boxes, three sets, weird problem

a,b,c

be three positive integers such that

a<b<c

. Consider the the sets

A,B,C

and

X

, defined as follows:

A=\{ 1,2,\ldots ,a \}

,

B=\{a+1,a+2,\ldots,b\}

,

C=\{b+1,b+2,\ldots ,c\}

and

X=A\cup B\cup C

. Determine, in terms of

a,b

and

c

, the number of ways of placing the elements of

X

in three boxes such that there are

x,y

and

z

elements in the first, second and third box respectively, knowing that: i)

x\le y\le z

; ii) elements of

B

cannot be put in the first box; iii) elements of

C

cannot be put in the third box.

Mulitple of n such that the sum of the digits of m is n

Prove that for all positive integers

n

, there exists a positive integer

m

which is a multiple of

n

and the sum of the digits of

m

is equal to

n

.

2

PA^2+PB^2=4PT^2

Let

C_1

be a circle with centre

O

, and let

AB

be a chord of the circle that is not a diameter.

M

is the midpoint of

AB

. Consider a point

T

on the circle

C_2

with diameter

OM

. The tangent to

C_2

at the point

T

intersects

C_1

at two points. Let

P

be one of these points. Show that

PA^2+PB^2=4PT^2

.

Number of connected subsets of {1,2, ... ,10}

A set of positive integers

X

is called connected if

|X|\ge 2

and there exist two distinct elements

m

and

n

of

X

such that

m

is a divisor of

n

. Determine the number of connected subsets of the set

\{1,2,\ldots,10\}

.

1

2

f(x+y) is at most f(x)+f(y), which is at most x+y

Determine all functions

f:\mathbb{R}\to\mathbb{R}

satisfying

f(x+y)\le f(x)+f(y)\le x+y

for all

x,y\in\mathbb{R}

.

Too easy to be in a mathematical competition surely...

Let

x

and

y

be two positive reals. Prove that

xy\le\frac{x^{n+2}+y^{n+2}}{x^n+y^n}

for all non-negative integers

n

.

2008 Pan African

Subcontests

Three boxes, three sets, weird problem

Mulitple of n such that the sum of the digits of m is n

PA^2+PB^2=4PT^2

Number of connected subsets of {1,2, ... ,10}

f(x+y) is at most f(x)+f(y), which is at most x+y

Too easy to be in a mathematical competition surely...