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Let n be a positive integer. Given a sequence \epsilon_1, ..., \epsilon_{n - 1} with \epsilon_i = 0 or \epsilon_i = 1 for each i = 1, ..., n - 1, the sequences a_0, ..., a_n and b_0, ..., b_n are constructed by the following rules: a_0 = b_0 = 1, a_1 = b_1 = 7,
a_{i + 1} = \left\{\begin{array}{cl}2a_{i - 1} + 3a_i\text{,} & \text{if } \epsilon_i = 0 \text{,}\\3a_{i - 1} + a_i\text{,} & \text{if } \epsilon_{i}= 1,\end{array}\right.
for each i = 1, ..., n - 1,
b_{i + 1} = \left\{\begin{array}{cl}2b_{i - 1} + 3b_i\text{,} & \text{if } \epsilon_{n - i} = 0 \text{,}\\3b_{i - 1} + b_i\text{,} & \text{if } \epsilon_{n - i} = 1 \text{,}\end{array}\right.
for each i = 1, ..., n - 1.

Prove that a_n = b_n.

Proposed by Ilya Bogdanov, Russia

Slični zadaci

Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighbouring buckets, empties them to the river and puts them back. Then the next round begins. The Stepmother goal's is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow?

Proposed by Gerhard Woeginger, Netherlands
Consider 2009 cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two player, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins.
(a) Does the game necessarily end?
(b) Does there exist a winning strategy for the starting player?

Proposed by Michael Albert, Richard Guy, New Zealand
Let A_0 = (a_1,\dots,a_n) be a finite sequence of real numbers. For each k\geq 0, from the sequence A_k = (x_1,\dots,x_k) we construct a new sequence A_{k + 1} in the following way.
1. We choose a partition \{1,\dots,n\} = I\cup J, where I and J are two disjoint sets, such that the expression
\left|\sum_{i\in I}x_i - \sum_{j\in J}x_j\right|
attains the smallest value. (We allow I or J to be empty; in this case the corresponding sum is 0.) If there are several such partitions, one is chosen arbitrarily.
2. We set A_{k + 1} = (y_1,\dots,y_n) where y_i = x_i + 1 if i\in I, and y_i = x_i - 1 if i\in J.
Prove that for some k, the sequence A_k contains an element x such that |x|\geq\frac n2.

Author: Omid Hatami, Iran
Let n > 1 be an integer. Find all sequences a_1, a_2, \ldots a_{n^2 + n} satisfying the following conditions:
\text{ (a) } a_i \in \left\{0,1\right\} for all 1 \leq i \leq n^2 + n;
\text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} for all 0 \leq i \leq n^2 - n
Author: unknown author, Serbia
Consider a m\times n rectangular board consisting of mn unit squares. Two of its unit squares are called adjacent if they have a common edge, and a path is a sequence of unit squares in which any two consecutive squares are adjacent. Two parths are called non-intersecting if they don't share any common squares.

Each unit square of the rectangular board can be colored black or white. We speak of a coloring of the board if all its mn unit squares are colored.

Let N be the number of colorings of the board such that there exists at least one black path from the left edge of the board to its right edge. Let M be the number of colorings of the board for which there exist at least two non-intersecting black paths from the left edge of the board to its right edge.

Prove that N^{2}\geq M\cdot 2^{mn}.
Let S_n be the number of sequences (a_1, a_2, \ldots, a_n), where a_i \in \{0,1\}, in which no six consecutive blocks are equal. Prove that S_n \rightarrow \infty when n \rightarrow \infty.