olymmath

ZA-SAMO 2009

South African Mathematics Olympiad 2009.

Other problems

Problem 1

Determine the smallest integer n>1n > 1 with the property that n2(n−1)n^2(n - 1) is divisible by 2009.

Problem 2

Let ABCDABCD be a rectangle and EE the reflection of AA with respect to the diagonal BDBD . If EB=ECEB = EC , what is the ratio ADAB\frac{AD}{AB} ?

Problem 3

Ten girls, numbered from 1 to 10, sit at a round table, in a random order. Each girl then receives a new number, namely the sum of her own number and those of her two neighbours. Prove that some girl receives a new number greater than 17.

Problem 4

Let x1,x2,…,xnx_1,x_2,\dots,x_n be a finite sequence of real numbersm mwhere 0<xi<10<x_i<1 for all i=1,2,…,ni=1,2,\dots,n . Put P=x1x2⋯xnP=x_1x_2\cdots x_n , S=x1+x2+⋯+xnS=x_1+x_2+\cdots+x_n and T=1x1+1x2+⋯+1xnT=\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n} . Prove that T−S1−P>2.\frac{T-S}{1-P}>2.

Problem 5

A game is played on a board with an infinite row of holes labelled 0,1,2,…0, 1, 2, \dots . Initially, 20092009 pebbles are put into hole 11 ; the other holes are left empty. Now steps are performed according to the following scheme:

(i) At each step, two pebbles are removed from one of the holes (if possible), and one pebble is put into each of the neighbouring holes.

(ii) No pebbles are ever removed from hole 00 .

(iii) The game ends if there is no hole with a positive label that contains at least two pebbles.

Show that the game always terminates, and that the number of pebbles in hole 00 at the end of the game is independent of the specific sequence of steps. Determine this number.

Problem 6

Let AA denote the set of real numbers xx such that 0≤x<10\le x<1 . A function f:A→Rf:A\to \mathbb{R} has the properties:

(i) f(x)=2f(x2)f(x)=2f(\frac{x}{2}) for all x∈Ax\in A ;

(ii) f(x)=1−f(x−12)f(x)=1-f(x-\frac{1}{2}) if 12≤x<1\frac{1}{2}\le x<1 .

Prove that

(a) f(x)+f(1−x)≥23f(x)+f(1-x)\ge \frac{2}{3} if xx is rational and 0<x<10<x<1 .

(b) There are infinitely many odd positive integers qq such that equality holds in (a) when x=1qx=\frac{1}{q} .