olymmath

OMCC 2006

Centroamerican and Caribbean Mathematical Olympiad 2006.

Day 1

Problem 1

  • Day 1

For 0≤d≤90 \leq d \leq 9 , we define the numbers Sd=1+d+d2+⋯+d2006S_{d}=1+d+d^{2}+\cdots+d^{2006}Find the last digit of the number S0+S1+⋯+S9.S_{0}+S_{1}+\cdots+S_{9}.

Answer, solutions and source

Problem 2

  • Day 1

Let Γ\Gamma and Γ′\Gamma' be two congruent circles centered at OO and O′O' , respectively, and let AA be one of their two points of intersection. BB is a point on Γ\Gamma , CC is the second point of intersection of ABAB and Γ′\Gamma' , and DD is a point on Γ′\Gamma' such that OBDO′OBDO' is a parallelogram. Show that the length of CDCD does not depend on the position of BB .

Answer, solutions and source

Problem 3

  • Day 1

For every natural number nn we define f(n)=⌊n+n+12⌋f(n)=\left\lfloor n+\sqrt{n}+\frac{1}{2}\right\rfloor Show that for every integer k≥1k \geq 1 the equation f(f(n))−f(n)=kf(f(n))-f(n)=k has exactly 2k−12k-1 solutions.

Answer, solutions and source

Day 2

Problem 4

  • Day 2

The product of several distinct positive integers is divisible by 20062{2006}^{2} . Determine the minimum value the sum of such numbers can take.

Answer, solutions and source

Problem 5

  • Day 2

The Olimpia country is formed by nn islands. The most populated one is called Panacenter, and every island has a different number of inhabitants. We want to build bridges between these islands, which we'll be able to travel in both directions, under the following conditions:

a) No pair of islands is joined by more than one bridge.

b) Using the bridges we can reach every island from Panacenter.

c) If we want to travel from Panacenter to every other island, in such a way that we use each bridge at most once, the number of inhabitants of the islands we visit is strictly decreasing.

Determine the number of ways we can build the bridges.

Answer, solutions and source

Problem 6

  • Day 2

Let ABCDABCD be a convex quadrilateral. I=AC∩BDI=AC\cap BD , and EE , HH , FF and GG are points on ABAB , BCBC , CDCD and DADA respectively, such that EF∩GH=IEF \cap GH= I . If M=EG∩ACM=EG \cap AC , N=HF∩ACN=HF \cap AC , show that AMIM⋅INCN=IAIC.\frac{AM}{IM}\cdot \frac{IN}{CN}=\frac{IA}{IC}.

Answer, solutions and source