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Neka je skup prirodnih brojeva podijeljen u intervale na sljedeći način:
U prvom intervalu je broj 1, u drugom brojevi 2 i 3, u trećem 4, 5 i 6 i u svakom idućem jedan broj više nego u prethodnom (brojevi u intervalima su uzastopni).
Neka je p_i udio prostih brojeva u i-tom intervalu.

a) Dokaži ili opovrgni: Postoji beskonačno brojeva k za koje je  p_{k+1} < p_k.

b) Dokaži ili opovrgni: Postoji beskonačno brojeva k za koje je  p_{k+1} > p_k.

Slični zadaci

Niz \{ a_n \} je zadan na ovaj način:
 a_0=0, \ a_1=1, \ a_n=2a_{n-1}+a_{n-2}, \ n>1.
Dokažite da 2^k dijeli a_n ako i samo ako 2^k dijeli n.
A positive integer N is called balanced, if N=1 or if N can be written as a product of an even number of not necessarily distinct primes. Given positive integers a and b, consider the polynomial P defined by P\!\left(x\right) = \left(x+a\right)\left(x+b\right).
a) Prove that there exist distinct positive integers a and b such that all the number P\!\left(1\right), P\!\left(2\right), ..., P\!\left(50\right) are balanced.
b) Prove that if P\!\left(n\right) is balanced for all positive integers n, then a=b.

Proposed by Jorge Tipe, Peru
Let a_1, a_2, \ldots, a_n be distinct positive integers, n\ge 3. Prove that there exist distinct indices i and j such that a_i + a_j does not divide any of the numbers 3a_1, 3a_2, \ldots, 3a_n.

Proposed by Mohsen Jamaali, Iran
Let n be a positive integer and let p be a prime number. Prove that if a, b, c are integers (not necessarily positive) satisfying the equations
a^n + pb = b^n + pc = c^n + pa
then a = b = c.

Proposed by Angelo Di Pasquale, Australia
Let b,n > 1 be integers. Suppose that for each k > 1 there exists an integer a_k such that b - a^n_k is divisible by k. Prove that b = A^n for some integer A.

Author: unknown author, Canada
Let m be a fixed integer greater than 1. The sequence x_0, x_1, x_2, \ldots is defined as follows:

x_i=  2^i if 0 \leq i\leq m-1 and x_i = \sum_{j=1}^{m}x_{i-j}, if i\geq m.

Find the greatest k for which the sequence contains k consecutive terms divisible by m.