MathDB

6

Part of 2003 IMO Shortlist

Problems(2)

Sequence inequality

Source: IMO ShortList 2003, algebra problem 6

6/9/2004
Let nn be a positive integer and let (x1,…,xn)(x_1,\ldots,x_n), (y1,…,yn)(y_1,\ldots,y_n) be two sequences of positive real numbers. Suppose (z2,…,z2n)(z_2,\ldots,z_{2n}) is a sequence of positive real numbers such that zi+j2≥xiyjz_{i+j}^2 \geq x_iy_j for all 1≤i,j≤n1\le i,j \leq n.
Let M=max⁡{z2,…,z2n}M=\max\{z_2,\ldots,z_{2n}\}. Prove that (M+z2+⋯+z2n2n)2≥(x1+⋯+xnn)(y1+⋯+ynn). \left( \frac{M+z_2+\dots+z_{2n}}{2n} \right)^2 \ge \left( \frac{x_1+\dots+x_n}{n} \right) \left( \frac{y_1+\dots+y_n}{n} \right).
[hide="comment"] Edited by Orl.
Proposed by Reid Barton, USA
inequalitiesfunctioncalculusIMO Shortlist
Nice and hard "nt" problem on digital representati

Source: German TST 2004, IMO ShortList 2003, combinatorics problem 6

7/15/2004
Let f(k)f(k) be the number of integers nn satisfying the following conditions:
(i) 0≤n<10k0\leq n < 10^k so nn has exactly kk digits (in decimal notation), with leading zeroes allowed;
(ii) the digits of nn can be permuted in such a way that they yield an integer divisible by 1111.
Prove that f(2m)=10f(2m−1)f(2m) = 10f(2m-1) for every positive integer mm.
Proposed by Dirk Laurie, South Africa
modular arithmeticcombinatoricsdecimal representationcountingfunctionIMO Shortlist