MathDB

4

Part of 2007 IMO Shortlist

Problems(3)

2007 SL A4

Source: ISL 2007, A4, Ukrainian TST 2008 Problem 5

6/24/2008
Find all functions f: \mathbb{R}^{ \plus{} }\to\mathbb{R}^{ \plus{} } satisfying f\left(x \plus{} f\left(y\right)\right) \equal{} f\left(x \plus{} y\right) \plus{} f\left(y\right) for all pairs of positive reals x x and y y. Here, \mathbb{R}^{ \plus{} } denotes the set of all positive reals.
Proposed by Paisan Nakmahachalasint, Thailand
algebrafunctional equationIMO Shortlist
Minumum difference of sums

Source: ISL 2007, C4, AIMO 2008, TST 3, P1

7/13/2008
Let A_0 \equal{} (a_1,\dots,a_n) be a finite sequence of real numbers. For each k≥0 k\geq 0, from the sequence A_k \equal{} (x_1,\dots,x_k) we construct a new sequence A_{k \plus{} 1} in the following way. 1. We choose a partition \{1,\dots,n\} \equal{} I\cup J, where I I and J J are two disjoint sets, such that the expression \left|\sum_{i\in I}x_i \minus{} \sum_{j\in J}x_j\right| attains the smallest value. (We allow I I or J 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 \plus{} 1} \equal{} (y_1,\dots,y_n) where y_i \equal{} x_i \plus{} 1 if i∈I i\in I, and y_i \equal{} x_i \minus{} 1 if i∈J i\in J. Prove that for some k k, the sequence Ak A_k contains an element x x such that ∣x∣≥n2 |x|\geq\frac n2. Author: Omid Hatami, Iran
combinatoricsalgorithmInequalitypartitionIMO Shortlist
2^{3k} divides difference of binomials but not 2^{3k+1}

Source: IMO Shortlist 2007, N4, AIMO 2008, TST 6, P2

7/13/2008
For every integer k≥2, k \geq 2, prove that 23k 2^{3k} divides the number \binom{2^{k \plus{} 1}}{2^{k}} \minus{} \binom{2^{k}}{2^{k \minus{} 1}} but 2^{3k \plus{} 1} does not.
Author: Waldemar Pompe, Poland
binomial coefficientsnumber theoryDivisibilityIMO ShortlistPolandHi