MathDB

3

Part of 1998 Romania Team Selection Test

Problems(4)

Partition such that there exists a,b,c with a^b=c

Source: 0

4/23/2011
Let m≥2m\ge 2 be an integer. Find the smallest positive integer n>mn>m such that for any partition with two classes of the set {m,m+1,…,n}\{ m,m+1,\ldots ,n \} at least one of these classes contains three numbers a,b,ca,b,c (not necessarily different) such that ab=ca^b=c.
Ciprian Manolescu
number theory proposednumber theory
f(x)=(x^2+x)^{2^n}+1 is irreducible

Source: Romanian TST 1998

4/23/2011
Show that for any positive integer nn the polynomial f(x)=(x2+x)2n+1f(x)=(x^2+x)^{2^n}+1 cannot be decomposed into the product of two integer non-constant polynomials.
Marius Cavachi
algebrapolynomialIrreducible
Find number of elements in the set A_n(k)

Source: Romanian TST 1998

4/23/2011
Let nn be a positive integer and Pn\mathcal{P}_n be the set of integer polynomials of the form a0+a1x+…+anxna_0+a_1x+\ldots +a_nx^n where ∣ai∣≤2|a_i|\le 2 for i=0,1,…,ni=0,1,\ldots ,n. Find, for each positive integer kk, the number of elements of the set An(k)={f(k)∣f∈Pn}A_n(k)=\{f(k)|f\in \mathcal{P}_n \}.
Marian Andronache
algebrapolynomialalgebra proposed
Numbers on each of the cases equal sum of neighbours iff..

Source: Romanian TST 1998

4/23/2011
The lateral surface of a cylinder of revolution is divided by n−1n-1 planes parallel to the base and mm parallel generators into mnmn cases (n≥1,m≥3)( n\ge 1,m\ge 3). Two cases will be called neighbouring cases if they have a common side. Prove that it is possible to write a real number in each case such that each number is equal to the sum of the numbers of the neighbouring cases and not all the numbers are zero if and only if there exist integers k,lk,l such that n+1n+1 does not divide kk and cos⁡2lπm+cos⁡kπn+1=12 \cos \frac{2l\pi}{m}+\cos\frac{k\pi}{n+1}=\frac{1}{2}
Ciprian Manolescu
trigonometryalgebra proposedalgebraPolynomials